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All questions of Unit 7: Modern Physics for Grade 9 Exam
Nucleus ”a” contains 5 protons and 5 neutrons and has radius R. The radius of nucleus ”b”, which contains 35 protons and 45 neutrons, is closest to:- a)2R
- b)8R
- c)1.4R
- d)R
Correct answer is option 'A'. Can you explain this answer?
Nucleus ”a” contains 5 protons and 5 neutrons and has radius R. The radius of nucleus ”b”, which contains 35 protons and 45 neutrons, is closest to:
a)
2R
b)
8R
c)
1.4R
d)
R
|
Gaurav Kumar answered |
R∝A1/3
A(mass no.)=n+p
R/x=(10/80)1/3
R/x=(13/23)1/3
R/x=1/2
X=2R
A(mass no.)=n+p
R/x=(10/80)1/3
R/x=(13/23)1/3
R/x=1/2
X=2R
Number of alpha particles N scattered at an angle θ during Rutherford’s alpha scattering experiment is :- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Number of alpha particles N scattered at an angle θ during Rutherford’s alpha scattering experiment is :
a)
b)
c)
d)
|
Sushil Kumar answered |
Answer :- a
Solution :- For a single scatterer, such as a single gold nucleus within a thin gold foil layer, the differential scattering cross section is defined as follows [2]:
dσ(θ, φ)/dΩ = flux scattered into element dΩ at angles (θ, φ)/incident flux per unit area
dσ/dΩ = (Iθ × A)/ (dΩ × I0 × NAvo × ρ × x(foil))........... (1)
where NAvo is Avogadro’s number, xfoil is the thickness of the target foil, A is the atomic mass of the material in the target foil, dΩ is the solid-angle of the detector, I0 is the unattenuated intensity of the alpha particle beam.
dσ/dΩ = [(ZZ0 e^2/4E )^2]/(1/sin^4(θ/2 ))
dσ/dΩ is directly proportional to 1/sin^4(θ/2 )
Select an incorrect alternative:
i. the radius of the nth orbit is proprtional to n2
ii. the total energy of the electron in the nth orbit is inversely proportional to n
iii. the angular momentum of the electron in nth orbit is an integral multiple of h/2π
iv. the magnitude of potential energy of the electron in any orbit is greater than its kinetic energy
- a)Statement i
- b)Statement iv
- c)Statement ii
- d)Statement iii
Correct answer is option 'C'. Can you explain this answer?
Select an incorrect alternative:
i. the radius of the nth orbit is proprtional to n2
ii. the total energy of the electron in the nth orbit is inversely proportional to n
iii. the angular momentum of the electron in nth orbit is an integral multiple of h/2π
iv. the magnitude of potential energy of the electron in any orbit is greater than its kinetic energy
i. the radius of the nth orbit is proprtional to n2
ii. the total energy of the electron in the nth orbit is inversely proportional to n
iii. the angular momentum of the electron in nth orbit is an integral multiple of h/2π
iv. the magnitude of potential energy of the electron in any orbit is greater than its kinetic energy
a)
Statement i
b)
Statement iv
c)
Statement ii
d)
Statement iii
|
Hansa Sharma answered |
Statement i. Radius of Bohr's orbit of hydrogen atom is given by
r= n2h2/4π2mKze2
or, r=(0.59A˚)(n2/z)
So, from expression we found r∝n2
Hence the 1st statement is correct.
Statement ii.
We know that
En=-13.6 x z2/n2
So, En ∝1/n2
Hence the 2nd statement is wrong.
Statement iii.Bohr defined these stable orbits in his second postulate. According to this postulate:
r= n2h2/4π2mKze2
or, r=(0.59A˚)(n2/z)
So, from expression we found r∝n2
Hence the 1st statement is correct.
Statement ii.
We know that
En=-13.6 x z2/n2
So, En ∝1/n2
Hence the 2nd statement is wrong.
Statement iii.Bohr defined these stable orbits in his second postulate. According to this postulate:
- An electron revolves around the nucleus in orbits
- The angular momentum of revolution is an integral multiple of h/2π – where Planck’s constant [h = 6.6 x 10-34 J-s].
- Hence, the angular momentum (L) of the orbiting electron is: L = nh/2 π
Hence the 3rd statement is correct.
Statement iv.According to Bohr's theory
Angular momentum of electron in an orbit will be Integral multiple of (h/2π)
Magnitude of potential energy is twice of kinetic energy of electron in an orbit
∣P.E∣=2∣K.E∣
K.E=(13.6ev)( z2/n2)
Hence, The 4th statement is correct.
Statement iv.According to Bohr's theory
Angular momentum of electron in an orbit will be Integral multiple of (h/2π)
Magnitude of potential energy is twice of kinetic energy of electron in an orbit
∣P.E∣=2∣K.E∣
K.E=(13.6ev)( z2/n2)
Hence, The 4th statement is correct.
What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 100 volts?a)12.3 nmb)0.127 nmc)123 nmd)1.23 nmCorrect answer is option 'B'. Can you explain this answer?
|
|
Geetika Shah answered |
and as we know 1Å = 0.1 nm, so 1.227Å = 0.122 nm
A sample of radioactive material contains 1018 atoms. The half life of the material is 2 days, then the activity of the sample is- a)3.5 x 1014 Bq
- b)3.5 x 1012 Bq
- c)7 x 1011 Bq
- d)7 x 1016 Bq
Correct answer is option 'B'. Can you explain this answer?
A sample of radioactive material contains 1018 atoms. The half life of the material is 2 days, then the activity of the sample is
a)
3.5 x 1014 Bq
b)
3.5 x 1012 Bq
c)
7 x 1011 Bq
d)
7 x 1016 Bq
|
Jyoti Sengupta answered |
To find activity of the sample --->which is the rate of disintegration.
Since radioactivity comes under 1o kinetics.
[R]=k[A] [A]-->amount of initial sample 1018 atoms
Given,
Half-life=2days
K=0.693/2x24x60x60 sec
R=(0.693/2x24x60x60)x1018
R≈3.5x1012 Bq
Rutherford’s experiments on scattering of alpha particles proved that:- a)atoms contain electrons
- b)number of positive charges is equal to the number of negative charges
- c)atom is mostly empty
- d)positive charge is uniformly distributed in the atom
Correct answer is option 'C'. Can you explain this answer?
Rutherford’s experiments on scattering of alpha particles proved that:
a)
atoms contain electrons
b)
number of positive charges is equal to the number of negative charges
c)
atom is mostly empty
d)
positive charge is uniformly distributed in the atom
|
Divey Sethi answered |
Most of the α-particle passed through the foil straight without suffering any change in their direction. This shows that most of the space inside the atom is empty or hollow.
A small fraction of α-particles was deflected through small angles and a few through larger angles. For this to happen α- particles (positively charged) must approach a heavy positively charged core inside the atom (like charges repel each other). This heavy positively charged core inside the atom was named as the nucleus.
A small fraction of α-particles was deflected through small angles and a few through larger angles. For this to happen α- particles (positively charged) must approach a heavy positively charged core inside the atom (like charges repel each other). This heavy positively charged core inside the atom was named as the nucleus.
A radioactive material decays by simultaneous emission of two particles with respective half lives 1620 and 810 years. The time in years, after which one fourth of the material remains is- a)4860
- b)2340
- c)1080.0
- d)3240
Correct answer is option 'C'. Can you explain this answer?
A radioactive material decays by simultaneous emission of two particles with respective half lives 1620 and 810 years. The time in years, after which one fourth of the material remains is
a)
4860
b)
2340
c)
1080.0
d)
3240
|
Nikita Singh answered |
Since, from Rutherford-Soddy law, the number of atoms left after half-lives is given by
N=N0(1/2)n
where, N0 is the original number of atoms.
The number of half-lives, n= time of decay/effective half−life
Relation between effective disintegration constant (λ) and half-life (T)
λ=ln2/T
∴λ1+λ2= (ln2/ T1)+ (ln2/ T2)
Effective half-life,
1/T=1/T1+1/T2=(1/1620)+(1/810)
1/T=1+2/1620 ⇒T=540yr
∴n=T/540
∴N=N0(1/2)t/540⇒N/N0=(1/2)2=(1/2)t/540
⇒t/540=2⇒t=2×540=1080yr
N=N0(1/2)n
where, N0 is the original number of atoms.
The number of half-lives, n= time of decay/effective half−life
Relation between effective disintegration constant (λ) and half-life (T)
λ=ln2/T
∴λ1+λ2= (ln2/ T1)+ (ln2/ T2)
Effective half-life,
1/T=1/T1+1/T2=(1/1620)+(1/810)
1/T=1+2/1620 ⇒T=540yr
∴n=T/540
∴N=N0(1/2)t/540⇒N/N0=(1/2)2=(1/2)t/540
⇒t/540=2⇒t=2×540=1080yr
The number of electrons in an atom X of atomic number Z and mass number A is- a)Zero
- b)A
- c)Z
- d)A-Z
Correct answer is option 'C'. Can you explain this answer?
The number of electrons in an atom X of atomic number Z and mass number A is
a)
Zero
b)
A
c)
Z
d)
A-Z
|
|
Sushil Kumar answered |
No of neutrons are given by: (A−Z)
Given an atomic number (Z) and mass number (A), you can find the number of protons, neutrons, and electrons in a neutral atom. For example, a lithium atom (Z=3,A=7 amu) contains three protons (found from Z), three electrons (as the number of protons is equal to the number of electrons in an atom), and four neutrons (7–3=4).
Given an atomic number (Z) and mass number (A), you can find the number of protons, neutrons, and electrons in a neutral atom. For example, a lithium atom (Z=3,A=7 amu) contains three protons (found from Z), three electrons (as the number of protons is equal to the number of electrons in an atom), and four neutrons (7–3=4).
Number of ejected photoelectrons increases with increase- a)never
- b)in frequency of light
- c)in wavelength of light
- d)in intensity of light
Correct answer is option 'D'. Can you explain this answer?
Number of ejected photoelectrons increases with increase
a)
never
b)
in frequency of light
c)
in wavelength of light
d)
in intensity of light
|
Rohan Singh answered |
A photon is the smallest possible quantum of light. In general when you turn up the intensity of light you are increasing the number of photons per second that are emitted by the light source. Therefore the intensity of the light can indeed be changed independently of the frequency (or color) of the light.
Cadmium rods are used in a nuclear reactor for- a)absorbing neutrons
- b)speeding up slow neutrons
- c)regulating the power level of the reactor.
- d)slowing down fast neutrons
Correct answer is option 'A'. Can you explain this answer?
Cadmium rods are used in a nuclear reactor for
a)
absorbing neutrons
b)
speeding up slow neutrons
c)
regulating the power level of the reactor.
d)
slowing down fast neutrons
|
Awantika Gupta answered |
Cadmium and boron rod both are used in cotroling the reactivity of uranium means slow down the rate of fission...
The energy of the incident photon is 20 eV and the work function of the photosensitive metal is 10 eV. What is the stopping potential?- a)30 V
- b)5 V
- c)10 V
- d)15 V
Correct answer is option 'C'. Can you explain this answer?
The energy of the incident photon is 20 eV and the work function of the photosensitive metal is 10 eV. What is the stopping potential?
a)
30 V
b)
5 V
c)
10 V
d)
15 V
|
Krishna Iyer answered |
Stopping potential (Vo) is given by
Vo=W/q where W is the work function and q are the charge of an electron.
Given W=20eV−10eV=10eV. Also, q=e
Hence, Vo=(10eV)/e=10V
Vo=W/q where W is the work function and q are the charge of an electron.
Given W=20eV−10eV=10eV. Also, q=e
Hence, Vo=(10eV)/e=10V
In order to increase the kinetic energy of ejected photoelectrons, there should be an increase in- a)wavelength of radiation
- b)intensity of radiation
- c)frequency of radiation
- d)Both the wavelength and intensity of radiation
Correct answer is option 'C'. Can you explain this answer?
In order to increase the kinetic energy of ejected photoelectrons, there should be an increase in
a)
wavelength of radiation
b)
intensity of radiation
c)
frequency of radiation
d)
Both the wavelength and intensity of radiation
|
Arjun Singhania answered |
The kinetic energy of emitted photoelectrons should increase with the light amplitude. The rate of electron emission, which is proportional to the measured electric current, should increase as the light frequency is increased.
All nuclides with same mass number A are called- a)isobars
- b)isoclines
- c)isotones
- d)isotopes
Correct answer is option 'A'. Can you explain this answer?
All nuclides with same mass number A are called
a)
isobars
b)
isoclines
c)
isotones
d)
isotopes
|
|
Rocky Handsome answered |
Isobars are atoms of different elements with the same mass number but different atomic numbers.
• Isotones are atomic nuclei with the same number of neutrons (N) and different number of protons(Z)
• Isotones are atomic nuclei with the same number of neutrons (N) and different number of protons(Z)
In Rutherford’s experiment, a thin gold foil was bombarded with alpha particles. According to Thomson’s “plum-pudding” model of the atom, what should have happened?- a)All the alpha particles would have been deflected by the foil.
- b)All the alpha particles should have bounced straight back from the foil.
- c)Alpha particles should have passed through the foil with little or no deflection.
- d)Alpha particles should have become embedded in the foil.
Correct answer is option 'C'. Can you explain this answer?
In Rutherford’s experiment, a thin gold foil was bombarded with alpha particles. According to Thomson’s “plum-pudding” model of the atom, what should have happened?
a)
All the alpha particles would have been deflected by the foil.
b)
All the alpha particles should have bounced straight back from the foil.
c)
Alpha particles should have passed through the foil with little or no deflection.
d)
Alpha particles should have become embedded in the foil.
|
Snehal Gosavi answered |
Correct Option C ===>ΔΔΔ
ΔAlfa particles are massive particles and they have speed when they bombarded....
ΔAccording plum pudding model if atom given by Thomson protons and electrons are equally distributed....
Δthat's why massive alpha particles will pass through the foil with little deflection (due to protons)......
ΔBcoz here protons are like spreaded cloud and not in nucleus ( massive part or atom) ....
Δand alpha particle is more massive than single proton....so they will not deflect due to protons and will pass through foil....
In hydrogen atom the kinetic energy of electron in an orbit of radius r is given by- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
In hydrogen atom the kinetic energy of electron in an orbit of radius r is given by
a)
b)
c)
d)
|
Swati Verma answered |
K.E. of nth orbit
=> (1/k) Ze2/2r
For H atom,
K.E.=(1/4πε) x (e2/2r)
=> (1/k) Ze2/2r
For H atom,
K.E.=(1/4πε) x (e2/2r)
90% of a radioactive sample is left undisintegrated after time τ has elapsed, what percentage of initial sample will decay in a total time2τ?- a)9%
- b)38%
- c)19%
- d)62%
Correct answer is option 'C'. Can you explain this answer?
90% of a radioactive sample is left undisintegrated after time τ has elapsed, what percentage of initial sample will decay in a total time2τ?
a)
9%
b)
38%
c)
19%
d)
62%
|
|
Krishna Iyer answered |
Given that 90% is left un-decayed after time 't'.
Hence, 10% decays in time 't'.
Initially assume that the amount of substance is 'x'
After time 't' 10% is decayed.
i.e. Amount of substance left =0.9x
After further time 't' another 10% is decayed.
i.e. 0.1×0.9x is decayed
Leaving behind 0.81x.
Hence after time 2t we see that 0.19x has decayed, which is 19%.
Hence, 10% decays in time 't'.
Initially assume that the amount of substance is 'x'
After time 't' 10% is decayed.
i.e. Amount of substance left =0.9x
After further time 't' another 10% is decayed.
i.e. 0.1×0.9x is decayed
Leaving behind 0.81x.
Hence after time 2t we see that 0.19x has decayed, which is 19%.
The nuclide 92U238 has all the following except- a)92 protons
- b)146 neutrons
- c)238 nucleons
- d)92 neutrons
Correct answer is option 'D'. Can you explain this answer?
The nuclide 92U238 has all the following except
a)
92 protons
b)
146 neutrons
c)
238 nucleons
d)
92 neutrons
|
|
Sushil Kumar answered |
The nuclide (nucleus) consists of neutrons and protons (when combined called nucleons).
Thus,
No. of protons in 92U238 = 92,
No. of neutrons = 146 (238 – 92)
No. of nucleons = 238 (146 + 92)
Thus,
No. of protons in 92U238 = 92,
No. of neutrons = 146 (238 – 92)
No. of nucleons = 238 (146 + 92)
Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material .If the frequency is halved and the intensity is doubled, the photoelectric current becomes- a)Doubled
- b)Quadrupled
- c)Halved
- d)zero
Correct answer is option 'D'. Can you explain this answer?
Light of frequency 1.5 times the threshold frequency is incident on a photosensitive material .If the frequency is halved and the intensity is doubled, the photoelectric current becomes
a)
Doubled
b)
Quadrupled
c)
Halved
d)
zero
|
Ayush Joshi answered |
If the frequency is halved and intensity is doubled, the frequency of incident light will become 15/2 = 0.75 times the threshold frequency. So, as ν<νo Hence, photoelectric current will be zero.
Photons of energy 6 eV are incident on a potassium surface of a work function 2.1 eV. What is the stopping potential?- a)-3.9V
- b)-8.1V
- c)-5V
- d)-1.9V
Correct answer is 'A'. Can you explain this answer?
Photons of energy 6 eV are incident on a potassium surface of a work function 2.1 eV. What is the stopping potential?
a)
-3.9V
b)
-8.1V
c)
-5V
d)
-1.9V
|
Priyanka Sharma answered |
From photo-electric equation, eV0= E−φ
eV0=(6−2.1)eV
V0= 3.9 V
stopping potential is a negative potential to stop e- at saturated current .
eV0=(6−2.1)eV
V0= 3.9 V
stopping potential is a negative potential to stop e- at saturated current .
The work function of a photoelectric material is 3.32 eV. The threshold frequency will be equal to- a)8 ×1014 HZ
- b)9 ×1014 HZ
- c)7 ×1014 HZ
- d)6 ×1014 HZ
Correct answer is option 'A'. Can you explain this answer?
The work function of a photoelectric material is 3.32 eV. The threshold frequency will be equal to
a)
8 ×1014 HZ
b)
9 ×1014 HZ
c)
7 ×1014 HZ
d)
6 ×1014 HZ
|
|
Anu Sharma answered |
The threshold frequency can be calculated using the formula:
threshold frequency = work function / Planck's constant
Given that the work function is 3.32 eV, we need to convert it to joules by multiplying it by the conversion factor 1.602 x 10^-19 J/eV:
work function = 3.32 eV * 1.602 x 10^-19 J/eV = 5.31264 x 10^-19 J
The value of Planck's constant is 6.626 x 10^-34 J·s.
Therefore, the threshold frequency is:
threshold frequency = 5.31264 x 10^-19 J / (6.626 x 10^-34 J·s)
threshold frequency ≈ 8.03 x 10^14 Hz
So, the threshold frequency will be approximately 8.03 x 10^14 Hz.
threshold frequency = work function / Planck's constant
Given that the work function is 3.32 eV, we need to convert it to joules by multiplying it by the conversion factor 1.602 x 10^-19 J/eV:
work function = 3.32 eV * 1.602 x 10^-19 J/eV = 5.31264 x 10^-19 J
The value of Planck's constant is 6.626 x 10^-34 J·s.
Therefore, the threshold frequency is:
threshold frequency = 5.31264 x 10^-19 J / (6.626 x 10^-34 J·s)
threshold frequency ≈ 8.03 x 10^14 Hz
So, the threshold frequency will be approximately 8.03 x 10^14 Hz.
We know that the Rutherford model of the atom is superior to the Thompson model because when alpha particles are scattered from atoms:- a)the deflected angle is always large
- b)they are usually observed with kinetic energy between 5 and 10 MeV
- c)some alpha particles are deflected to large angles
- d)the deflected angle is usually small
Correct answer is option 'C'. Can you explain this answer?
We know that the Rutherford model of the atom is superior to the Thompson model because when alpha particles are scattered from atoms:
a)
the deflected angle is always large
b)
they are usually observed with kinetic energy between 5 and 10 MeV
c)
some alpha particles are deflected to large angles
d)
the deflected angle is usually small
|
Tanuja Kapoor answered |
In ruther ford experiment he suggest that all the positive charge and mass are concentrated at the centre when he bombarded the alpha partical which is dipositive in nature and when it is more close to centre it get deflect to a large angle and with increase of closenes to centre its deflection angle increase and some alpha partical deflect to 180 degree so it prove that all the positive charge and mass are concentrated at the centre where as acccording to thomson atom is hard solid sphere in which its total +ve charge and mass uniformalyy distributed on the surface and electrone reside as seed in watermelon ( plum pudding model)
A photo-sensitive material would emit electrons, if excited by photons beyond a threshold. To overcome the threshold, one would increase the:- a) voltage applied to the light source
- b) intensity of light
- c) wavelength of light
- d) frequency of light
Correct answer is option 'D'. Can you explain this answer?
A photo-sensitive material would emit electrons, if excited by photons beyond a threshold. To overcome the threshold, one would increase the:
a)
voltage applied to the light source
b)
intensity of light
c)
wavelength of light
d)
frequency of light
|
|
Rohan Singh answered |
The emission of photoelectron takes place only, when the frequency of the incident light is above a certain critical value, characteristic of that metal. The critical value of frequency is known as the threshold frequency for the metal of the emitting electrode.
Suppose that when light of certain frequency is incident over a metal surface, the photo- electrons are emitted. To take photoelectric current zero, a particular value of stopping potential will be needed. If we go on reducing the frequency of incident light, the value of stopping potential will also go on decreasing. At certain value of frequency v0, the photoelectric current will become zero, even when no retarding potential is applied. This frequency vq corresponds to the threshold for the metal surface. The emission of photoelectrons does not take place, till frequency of incident light is below this value.
α-rays are- a)helium nuclei
- b)heavy nuclei
- c)lithium nuclei
- d)hydrogen nuclei
Correct answer is option 'A'. Can you explain this answer?
α-rays are
a)
helium nuclei
b)
heavy nuclei
c)
lithium nuclei
d)
hydrogen nuclei
|
|
Ræjû Bhæï answered |
Alpha particles, also called alpha rays or alpha radiation, consist of two protons and two neutrons bound together into a particle identical to a helium-4 nucleus. They are generally produced in the process of alpha decay, but may also be produced in other ways.
Can you explain the answer of this question below:Light from a bulb is falling on a wooden table but no photo electrons are emitted as
- A:
Work function of wood is less
- B:
Work function of wood is more
- C:
It depends on the frequency
- D:
It is independent of work function
The answer is b.
Light from a bulb is falling on a wooden table but no photo electrons are emitted as
Work function of wood is less
Work function of wood is more
It depends on the frequency
It is independent of work function
|
Rishika Patel answered |
Explanation:
When light falls on a metal surface, electrons may be emitted from the metal surface. This phenomenon is known as the photoelectric effect. The electrons emitted from the metal surface are called photoelectrons.
The photoelectric effect can be explained by considering that light is made up of photons. Each photon has a certain amount of energy, given by its frequency. When a photon strikes a metal surface, its energy can be transferred to an electron in the metal. If the energy of the photon is greater than the work function of the metal, the electron can be emitted from the metal surface.
In the case of a wooden table, the work function of the wood is more than the energy of the photons of the light falling on it. Therefore, no photoelectrons are emitted. This is because the energy of the photons is not enough to overcome the work function of the wood.
Key Points:
- The photoelectric effect is the emission of electrons from a metal surface when light falls on it.
- The energy of a photon is given by its frequency.
- If the energy of a photon is greater than the work function of the metal, electrons can be emitted from the metal surface.
- In the case of a wooden table, the work function of the wood is more than the energy of the photons of the light falling on it, so no photoelectrons are emitted.
When light falls on a metal surface, electrons may be emitted from the metal surface. This phenomenon is known as the photoelectric effect. The electrons emitted from the metal surface are called photoelectrons.
The photoelectric effect can be explained by considering that light is made up of photons. Each photon has a certain amount of energy, given by its frequency. When a photon strikes a metal surface, its energy can be transferred to an electron in the metal. If the energy of the photon is greater than the work function of the metal, the electron can be emitted from the metal surface.
In the case of a wooden table, the work function of the wood is more than the energy of the photons of the light falling on it. Therefore, no photoelectrons are emitted. This is because the energy of the photons is not enough to overcome the work function of the wood.
Key Points:
- The photoelectric effect is the emission of electrons from a metal surface when light falls on it.
- The energy of a photon is given by its frequency.
- If the energy of a photon is greater than the work function of the metal, electrons can be emitted from the metal surface.
- In the case of a wooden table, the work function of the wood is more than the energy of the photons of the light falling on it, so no photoelectrons are emitted.
In a photon-particle collision (such as photon-electron collision) the quantity which is not conserved is- a)total momentum
- b)number of photons
- c)total energy
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
In a photon-particle collision (such as photon-electron collision) the quantity which is not conserved is
a)
total momentum
b)
number of photons
c)
total energy
d)
None of the above
|
|
Rohan Singh answered |
Energy and momentum are conserved, resulting in a reduction of both for the scattered photon. ... This phenomenon could be handled as a collision between two particles—a photon and an electron at rest in the material. Energy and momentum are conserved in the collision.
Fluorescent lamps are more efficient than incandescent lamps in converting electrical energy to visible light because- a)they produce more white light
- b)they do not use uv radiations
- c)they do not waste as much energy producing (invisible) infrared photons
- d)they do not waste as much energy producing visible photons
Correct answer is option 'C'. Can you explain this answer?
Fluorescent lamps are more efficient than incandescent lamps in converting electrical energy to visible light because
a)
they produce more white light
b)
they do not use uv radiations
c)
they do not waste as much energy producing (invisible) infrared photons
d)
they do not waste as much energy producing visible photons
|
|
Riya Banerjee answered |
The phosphor fluoresces to produce light. A fluorescent bulb produces less heat, so it is much more efficient. This makes fluorescent bulbs four to six times more efficient than incandescent bulbs. That's why you can buy a 15-watt fluorescent bulb that produces the same amount of light as a 60-watt incandescent bulb.
Sodium surface is illuminated by ultraviolet and visible radiation successively and the stopping potential determined. The stopping potential is- a)Greater with visible light
- b)Equal in both cases
- c)Greater with ultraviolet light
- d)Infinite in both cases
Correct answer is 'C'. Can you explain this answer?
Sodium surface is illuminated by ultraviolet and visible radiation successively and the stopping potential determined. The stopping potential is
a)
Greater with visible light
b)
Equal in both cases
c)
Greater with ultraviolet light
d)
Infinite in both cases
|
|
Riya Banerjee answered |
λ for U.V is less than λ for visible light
ν for U.V is greater than ν for visible light
∴ potential is greater for U.V light.
as K.Eα1/λ
ν for U.V is greater than ν for visible light
∴ potential is greater for U.V light.
as K.Eα1/λ
At a given time there are 25% undecayed nuclei in a sample. After 10 seconds number of undecayed nuclei reduces to 12.5%. Then mean life of the nuclei will be about- a)22 sec
- b)10 sec
- c)12 sec
- d)15 sec
Correct answer is option 'D'. Can you explain this answer?
At a given time there are 25% undecayed nuclei in a sample. After 10 seconds number of undecayed nuclei reduces to 12.5%. Then mean life of the nuclei will be about
a)
22 sec
b)
10 sec
c)
12 sec
d)
15 sec
|
|
Lavanya Menon answered |
Half-life of radioactive sample, i.e., the time in which the number of undecayed nuclei becomes half (T) is 10 s.
Mean life, τ=T/loge2=10s/0.693=1.443×10=14.43s ≈ 15s
Mean life, τ=T/loge2=10s/0.693=1.443×10=14.43s ≈ 15s
The half life of radon is 3.8 days. After how many days will only one twentieth of a radon sample be left over?
a)10.00 daysb)5.45 daysc)15.45 daysd)16.45 daysCorrect answer is option 'D'. Can you explain this answer?
|
|
Suresh Iyer answered |
Let the initial amount of radon be N0 and the amount left after t days be N which is equal to N0/2
An α – particle and a deutron are accelerated through the same potential difference. What will be the ratio of their de-Broglie wavelength?- a)1 / 3
- b)1 / 5
- c)1 / 4
- d)1 / 2
Correct answer is 'D'. Can you explain this answer?
An α – particle and a deutron are accelerated through the same potential difference. What will be the ratio of their de-Broglie wavelength?
a)
1 / 3
b)
1 / 5
c)
1 / 4
d)
1 / 2
|
|
Abhay Iyer answered |
Mass of alpha(a)= 4.
mass of deutron(d) =2.
wavelength=h/mv
Potential are same of both particle so their speed will be same .
Here, wavelength is inversely proportional to mass
so,
wavelength of alpha/wavelength of deutron = mass of deutron /mass of alpha.= 2/4=1/2
mass of deutron(d) =2.
wavelength=h/mv
Potential are same of both particle so their speed will be same .
Here, wavelength is inversely proportional to mass
so,
wavelength of alpha/wavelength of deutron = mass of deutron /mass of alpha.= 2/4=1/2
The de-Broglie wavelength of an electron is 1.0 nm. What is the retarding potential required to stop it?- a)1.5 V
- b)3.5 V
- c)6.5 V
- d)4.5 V
Correct answer is option 'A'. Can you explain this answer?
The de-Broglie wavelength of an electron is 1.0 nm. What is the retarding potential required to stop it?
a)
1.5 V
b)
3.5 V
c)
6.5 V
d)
4.5 V
|
|
Neha Sharma answered |
λ=h/P
Here P= √2mk
K=kinetic energy
λ=h/√2mk
By energy conversion
K=eVs
λ=h/√2meVs
√Vs=h/ λ√2me
Vs=h2/ λ2me
Vs=(6.626x10-34)2/(1x10-9)2x2x9.11x10-31xe
Where e=1.602x10-19
Vs=43.9x10-68/29.1x10-68
=1.508v
Vs=1.5v
Here P= √2mk
K=kinetic energy
λ=h/√2mk
By energy conversion
K=eVs
λ=h/√2meVs
√Vs=h/ λ√2me
Vs=h2/ λ2me
Vs=(6.626x10-34)2/(1x10-9)2x2x9.11x10-31xe
Where e=1.602x10-19
Vs=43.9x10-68/29.1x10-68
=1.508v
Vs=1.5v
Which of these is true?- a)The alpha particles used in Rutherford’s experiment are positively charged particles
- b)Dalton assumed that atoms are made up of electrons, protons, and neutrons
- c)In Rutherford’s alpha scattering experiment, all of the alpha particles passed through the gold foil.
- d)JJ Thomson determined the charge and mass of electrons
Correct answer is 'A'. Can you explain this answer?
Which of these is true?
a)
The alpha particles used in Rutherford’s experiment are positively charged particles
b)
Dalton assumed that atoms are made up of electrons, protons, and neutrons
c)
In Rutherford’s alpha scattering experiment, all of the alpha particles passed through the gold foil.
d)
JJ Thomson determined the charge and mass of electrons
|
Shalini Basu answered |
**Explanation:**
The correct answer is **a) The alpha particles used in Rutherford's experiment are positively charged particles**.
Rutherford's experiment, also known as the gold foil experiment, was conducted in 1909 by Ernest Rutherford and his colleagues Hans Geiger and Ernest Marsden. The experiment aimed to understand the structure of the atom and investigate the distribution of positive charge within it.
In this experiment, Rutherford and his team bombarded a thin gold foil with a beam of alpha particles. Alpha particles are positively charged particles that consist of two protons and two neutrons, which are the same as helium nuclei. They are emitted from a radioactive source, such as radium or polonium.
Rutherford observed that while most of the alpha particles passed straight through the gold foil, some of them were deflected at different angles, and a very small fraction bounced back in the direction opposite to the source. This unexpected result led to the discovery of the atomic nucleus and the concept of a mostly empty space within the atom.
Based on the observations from the experiment, Rutherford proposed a new atomic model known as the nuclear model. According to this model, atoms have a dense, positively charged nucleus at the center, which contains most of the atom's mass. The electrons, which are negatively charged particles, orbit around the nucleus in specific energy levels.
Therefore, the correct answer is a) The alpha particles used in Rutherford's experiment are positively charged particles.
The correct answer is **a) The alpha particles used in Rutherford's experiment are positively charged particles**.
Rutherford's experiment, also known as the gold foil experiment, was conducted in 1909 by Ernest Rutherford and his colleagues Hans Geiger and Ernest Marsden. The experiment aimed to understand the structure of the atom and investigate the distribution of positive charge within it.
In this experiment, Rutherford and his team bombarded a thin gold foil with a beam of alpha particles. Alpha particles are positively charged particles that consist of two protons and two neutrons, which are the same as helium nuclei. They are emitted from a radioactive source, such as radium or polonium.
Rutherford observed that while most of the alpha particles passed straight through the gold foil, some of them were deflected at different angles, and a very small fraction bounced back in the direction opposite to the source. This unexpected result led to the discovery of the atomic nucleus and the concept of a mostly empty space within the atom.
Based on the observations from the experiment, Rutherford proposed a new atomic model known as the nuclear model. According to this model, atoms have a dense, positively charged nucleus at the center, which contains most of the atom's mass. The electrons, which are negatively charged particles, orbit around the nucleus in specific energy levels.
Therefore, the correct answer is a) The alpha particles used in Rutherford's experiment are positively charged particles.
In an experiment of photoelectric emission for incident light of 4000 A0, the stopping potential is 2V. If the wavelength of incident light is made 3000 A0, then stopping potential will be- a)zero
- b)more than 2 volt
- c)2 Volt
- d)less than 2 volt
Correct answer is option 'B'. Can you explain this answer?
In an experiment of photoelectric emission for incident light of 4000 A0, the stopping potential is 2V. If the wavelength of incident light is made 3000 A0, then stopping potential will be
a)
zero
b)
more than 2 volt
c)
2 Volt
d)
less than 2 volt
|
|
Geetika Shah answered |
The maximum kinetic energy for the photoelectrons is
Emax=hν−ϕ
where, ν is the frequency of incident light and ϕ is photoelectric work function of metal.
If Vo is the stopping potential then
eV0=h(c/λ)−ϕ .....................(since, ν=c/λ)
As per the problem, for incident light of 4000Ao, the stopping potential is 2V. When the wavelength of incident light is reduced to 3000Ao, then the stopping potential will increase to value more than 2V(as per the above equation).
Emax=hν−ϕ
where, ν is the frequency of incident light and ϕ is photoelectric work function of metal.
If Vo is the stopping potential then
eV0=h(c/λ)−ϕ .....................(since, ν=c/λ)
As per the problem, for incident light of 4000Ao, the stopping potential is 2V. When the wavelength of incident light is reduced to 3000Ao, then the stopping potential will increase to value more than 2V(as per the above equation).
In the following reaction
What is following value of a?- a)14
- b)10
- c)16
- d)12
Correct answer is option 'D'. Can you explain this answer?
In the following reaction
What is following value of a?
a)
14
b)
10
c)
16
d)
12
|
|
Tanuja Kapoor answered |
The sum of the atomic no. and atomic mass no. on the reactant and product should be equal .
therefore 9+4 = a+1
a = 12
therefore 9+4 = a+1
a = 12
The atomic number Z of the nucleus is- a)Number of deutrons.
- b)Number of neutrons in it.
- c)Number if electrons in it.
- d)Number of protons in it.
Correct answer is option 'D'. Can you explain this answer?
The atomic number Z of the nucleus is
a)
Number of deutrons.
b)
Number of neutrons in it.
c)
Number if electrons in it.
d)
Number of protons in it.
|
|
Shraddha Dey answered |
**Explanation:**
The atomic number (Z) of an atom refers to the number of protons in the nucleus. Here, we will discuss why the correct answer is option 'D' and explain the significance of atomic number in an atom.
**Atomic Number (Z):**
The atomic number of an atom is a fundamental property that determines its identity and place in the periodic table. It is denoted by the symbol 'Z'. Each element on the periodic table has a unique atomic number.
**Protons in the Nucleus:**
Protons are subatomic particles that carry a positive charge. They are located in the nucleus of an atom, which is the central core of the atom. The number of protons in the nucleus is equal to the atomic number of the atom.
**Electrons in the Atom:**
Electrons are subatomic particles that carry a negative charge. They orbit around the nucleus in specific energy levels or shells. The number of electrons in a neutral atom is equal to the number of protons, ensuring that the atom has a balanced charge overall.
**Neutrons in the Nucleus:**
Neutrons are subatomic particles that have no charge (they are electrically neutral). They are also located in the nucleus along with protons. The number of neutrons in an atom can vary, resulting in different isotopes of the same element. Isotopes have the same atomic number (same number of protons) but different mass numbers (different number of neutrons).
**Significance of Atomic Number:**
The atomic number is a crucial characteristic of an atom because it determines the element's identity. Elements are organized in increasing order of their atomic numbers on the periodic table. For example, hydrogen has an atomic number of 1, helium has an atomic number of 2, and so on.
The atomic number defines the unique properties and behavior of an element. It determines the number of electrons in the atom, which influences the atom's chemical reactivity and bonding. It also provides information about the element's position in the periodic table, its atomic mass, and its isotopes.
Therefore, the correct answer to the given question is option 'D' – the atomic number (Z) of the nucleus represents the number of protons in it.
The atomic number (Z) of an atom refers to the number of protons in the nucleus. Here, we will discuss why the correct answer is option 'D' and explain the significance of atomic number in an atom.
**Atomic Number (Z):**
The atomic number of an atom is a fundamental property that determines its identity and place in the periodic table. It is denoted by the symbol 'Z'. Each element on the periodic table has a unique atomic number.
**Protons in the Nucleus:**
Protons are subatomic particles that carry a positive charge. They are located in the nucleus of an atom, which is the central core of the atom. The number of protons in the nucleus is equal to the atomic number of the atom.
**Electrons in the Atom:**
Electrons are subatomic particles that carry a negative charge. They orbit around the nucleus in specific energy levels or shells. The number of electrons in a neutral atom is equal to the number of protons, ensuring that the atom has a balanced charge overall.
**Neutrons in the Nucleus:**
Neutrons are subatomic particles that have no charge (they are electrically neutral). They are also located in the nucleus along with protons. The number of neutrons in an atom can vary, resulting in different isotopes of the same element. Isotopes have the same atomic number (same number of protons) but different mass numbers (different number of neutrons).
**Significance of Atomic Number:**
The atomic number is a crucial characteristic of an atom because it determines the element's identity. Elements are organized in increasing order of their atomic numbers on the periodic table. For example, hydrogen has an atomic number of 1, helium has an atomic number of 2, and so on.
The atomic number defines the unique properties and behavior of an element. It determines the number of electrons in the atom, which influences the atom's chemical reactivity and bonding. It also provides information about the element's position in the periodic table, its atomic mass, and its isotopes.
Therefore, the correct answer to the given question is option 'D' – the atomic number (Z) of the nucleus represents the number of protons in it.
The radius of a nucleus is directly proportional to (A=mass number)- a)A1/3
- b)A2
- c)A3
- d)A1/2
Correct answer is option 'A'. Can you explain this answer?
The radius of a nucleus is directly proportional to (A=mass number)
a)
A1/3
b)
A2
c)
A3
d)
A1/2
|
|
Riya Banerjee answered |
For A nucleons
R=RoA1/3 [Ro=constant]
So, R∝A1/3
R=RoA1/3 [Ro=constant]
So, R∝A1/3
Light of wavelength 4000 Â is incident on a metal of work function 3.2 x 10-19 J. What is the maximum kinetic energy of the emitted electron?
- a)2.2 x 10-19 J
- b)1.75 x 10-19 J
- c)0.75 x 10-19 J
- d)1.1 x 10-19 J
Correct answer is option 'B'. Can you explain this answer?
Light of wavelength 4000 Â is incident on a metal of work function 3.2 x 10-19 J. What is the maximum kinetic energy of the emitted electron?
a)
2.2 x 10-19 J
b)
1.75 x 10-19 J
c)
0.75 x 10-19 J
d)
1.1 x 10-19 J
|
|
Om Desai answered |
Given,
λ= 4000 Å ;W=3.2× 10-19 j
as we know that
K.E = (hc/ λ) - W
= (6.6× 10-34 × 3 × 108/ 4000×10-10 )- 3.20× 10-19
= 1.75 × 10-19
λ= 4000 Å ;W=3.2× 10-19 j
as we know that
K.E = (hc/ λ) - W
= (6.6× 10-34 × 3 × 108/ 4000×10-10 )- 3.20× 10-19
= 1.75 × 10-19
Emission line spectra of different elements- a)is not different
- b)is the same if both elements are at the same temperature
- c)is the same if both elements are in liquid form
- d)is different
Correct answer is option 'D'. Can you explain this answer?
Emission line spectra of different elements
a)
is not different
b)
is the same if both elements are at the same temperature
c)
is the same if both elements are in liquid form
d)
is different
|
Adithya Shasan answered |
Option D : Emission line spectra of different elements is different..
What is the main source of energy of the sun?- a)Nuclear fission of heavier unstable elements in the sun
- b)Combustion of pure carbon present in the sun
- c)Gravitational energy liberated during the slow contraction of the sun.
- d)Nuclear fusion of lighter elements in the sun.
Correct answer is option 'D'. Can you explain this answer?
What is the main source of energy of the sun?
a)
Nuclear fission of heavier unstable elements in the sun
b)
Combustion of pure carbon present in the sun
c)
Gravitational energy liberated during the slow contraction of the sun.
d)
Nuclear fusion of lighter elements in the sun.
|
Kanika S answered |
Option d
When H and He in sun undergoes fusion large amount of heat is released.
When H and He in sun undergoes fusion large amount of heat is released.
The average number of neutrons released by the fission of one uranium atom is
a)3.0b)2c)2.5d)1Correct answer is option 'C'. Can you explain this answer?
|
|
Bhanu Saini answered |
Fission result in the production of typically 2 or 3 neutron so on the average about 2.5 neutron released per unit. so correct answer is option a
for option c one uranium atom split into one barium and one krypton atom releasing 3 neutron.
but in this question average is asking so according to me and books 2.5 is correct
for option c one uranium atom split into one barium and one krypton atom releasing 3 neutron.
but in this question average is asking so according to me and books 2.5 is correct
In hydrogen atom the angular momentum of the electron in the lowest energy state is- a)2h
- b)h/2π
- c)2π/h
- d)h/π
Correct answer is option 'B'. Can you explain this answer?
In hydrogen atom the angular momentum of the electron in the lowest energy state is
a)
2h
b)
h/2π
c)
2π/h
d)
h/π
|
|
Kiran Khanna answered |
C)h/π
d)h
The correct answer is d) h.
The angular momentum of an electron in the hydrogen atom is given by the formula L = nħ, where n is the principal quantum number and ħ is the reduced Planck constant.
In the lowest energy state, the electron is in the ground state with n = 1. Therefore, the angular momentum is L = 1ħ = h.
d)h
The correct answer is d) h.
The angular momentum of an electron in the hydrogen atom is given by the formula L = nħ, where n is the principal quantum number and ħ is the reduced Planck constant.
In the lowest energy state, the electron is in the ground state with n = 1. Therefore, the angular momentum is L = 1ħ = h.
Wavelength of light incident on a photo cell is 3000 Â, if stopping potential is 2.5 volts, then work function of the cathode of photo cell is- a)1.64 eV
- b)1.56 eV
- c)1.52 eV
- d)1.41 eV
Correct answer is option 'A'. Can you explain this answer?
Wavelength of light incident on a photo cell is 3000 Â, if stopping potential is 2.5 volts, then work function of the cathode of photo cell is
a)
1.64 eV
b)
1.56 eV
c)
1.52 eV
d)
1.41 eV
|
|
Neha Sharma answered |
The Stopping potential =2.5V.
or, Kinetic energy=2.5eV.
We know that,
Incident energy =work function + Kinetic energy.
To get incident energy in e.V,
We also know that,
The Incident energy =12400/λ Å
Incident energy=work function + kinetic energy.
12400/3000 = work function + 2.5e.v.
4.13-2.5 = work function
work function=1.64 e.V
or, Kinetic energy=2.5eV.
We know that,
Incident energy =work function + Kinetic energy.
To get incident energy in e.V,
We also know that,
The Incident energy =12400/λ Å
Incident energy=work function + kinetic energy.
12400/3000 = work function + 2.5e.v.
4.13-2.5 = work function
work function=1.64 e.V
In which case is electron emission from a metal not known?- a)Illuminating a metal surface with suitable light
- b)Applying a very strong magnetic field to a metal
- c)Applying a very strong electric field to a metal
- d)Heating the metal sufficiently
Correct answer is option 'B'. Can you explain this answer?
In which case is electron emission from a metal not known?
a)
Illuminating a metal surface with suitable light
b)
Applying a very strong magnetic field to a metal
c)
Applying a very strong electric field to a metal
d)
Heating the metal sufficiently
|
|
Divey Sethi answered |
The correct answer would be option B. Applying a very strong magnetic field.
Applying a very strong magnetic field to a metal we don’t know the electron emission.
Applying a very strong magnetic field to a metal we don’t know the electron emission.
Given h = 6.6 ×10−34 joule sec, the momentum of each photon in a given radiation is 3.3 ×10−29 kg metre/sec. The frequency of radiation is- a)1.6 ×1013Hz
- b)1.5 ×1013Hz
- c)1.7 ×1013Hz
- d)1.8 ×1013Hz
Correct answer is option 'B'. Can you explain this answer?
Given h = 6.6 ×10−34 joule sec, the momentum of each photon in a given radiation is 3.3 ×10−29 kg metre/sec. The frequency of radiation is
a)
1.6 ×1013Hz
b)
1.5 ×1013Hz
c)
1.7 ×1013Hz
d)
1.8 ×1013Hz
|
|
Lavanya Menon answered |
Frequency=C/ λ
λ=h/P
frequency=C P/ h
f=(3×108×3.3×10-29)/6.6×10-34
f=3× 1013/2
f=1.5×1013 Hz
λ=h/P
frequency=C P/ h
f=(3×108×3.3×10-29)/6.6×10-34
f=3× 1013/2
f=1.5×1013 Hz
The distance of closest approach when a 15.0 MeV proton approaches gold nucleus (Z = 79) is- a)758 fm
- b)7.58 fm
- c)75.8 fm
- d)0.758 fm
Correct answer is option 'B'. Can you explain this answer?
The distance of closest approach when a 15.0 MeV proton approaches gold nucleus (Z = 79) is
a)
758 fm
b)
7.58 fm
c)
75.8 fm
d)
0.758 fm
|
|
Riya Banerjee answered |
Correct Answer :- b
Explanation : E = 15.0MeV
= 15 * 106 eV
= 15 * 106 * 1.6 * 10-19 J
= 15 * 1.6 * 10-13 J
E = (1/4πεo)*(ze2/r02)
r0 = (1/4πεo)*(ze2/E)
r0 = (9*109*79*(1.6*10-19)2)/(15*1.6*10-13)
= 75.84 * 10-16 m
= 7.58 fm
Nuclear mass M is found to be- a)always greater than total mass of its individual protons and neutrons
- b)always equal to the total mass of its individual neutrons
- c)always equal to the total mass of its individual protons and neutrons
- d)always less than total mass of its individual protons and neutrons
Correct answer is option 'D'. Can you explain this answer?
Nuclear mass M is found to be
a)
always greater than total mass of its individual protons and neutrons
b)
always equal to the total mass of its individual neutrons
c)
always equal to the total mass of its individual protons and neutrons
d)
always less than total mass of its individual protons and neutrons
|
|
Ritu Singh answered |
The actual mass is always less than the sum of the individual masses of the constituent protons and neutrons because energy is removed when the nucleus is formed. This energy has mass, which is removed from the total mass of the original particles.
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