Dual Nature of Matter and Radiation Test

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Dual Nature of Matter and Radiation Test - 10

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Question 1
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Photons with energy 5eV are incident on a cathode C, on a photoelectric cell. The maximum energy of the emitted photoelectrons is 2eV. When photons of energy 6eV are incident on C, no photoelectrons will reach the anode A if the stopping potential of A relative to C is

A
3V

B
^_ 3V

C
^_ 1V

D
4V

Solution

When 5eV is incident the kinetic energy is 2eV it simply means the work function is W=5eV −2eV=3eV
Similarly, when 6eV is incident the kinetic energy should be 6eV −W=6eV −3eV=3eV
it simply means to stop them we need a negative potential at anode equal to 3eV/e =3V
So, the answer is −3V i.e. option B is correct.
Question 2
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By increasing the intensity of incident light keeping frequency (v >v₀ ) fixed on the surface of metal

A
kinetic energy of the photoelectrons increases

B
number of emitted electrons increases

C
kinetic energy and number of electrons increases

D
no effect

Solution

The number of photoelectrons emitted per second from a photosensitive plate is directly proportional to the intensity of the incident radiation. ... For the same frequency of light and increased intensity, the saturation current is found to increase, but the cut-off potential is found to remain constant.
The number of electrons also changes because of the probability that each photon results in an emitted electron are a function of photon energy. If the intensity of the incident radiation of a given frequency is increased, there is no effect on the kinetic energy of each photo electron.
Question 3
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In a photoelectric experiment, electrons are ejected from metals X and Y by light of intensity I and frequency f. The potential difference V required to stop the electrons is measured for various frequencies. If Y has a greater work function than X ; which one of the following graphs best illustrates the expected results ?

A

B

C

D

Solution

The answer is option A.
First, the gradient of the graph cannot change (always = h/e ), so answers are A or D.
If Y has a greater work function than X, then the graph for Y should have a more negative y-intercept. therefore figure in option A depicts the concept.
Question 4
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A image of the sun of formed by a lens of focal-length of 30 cm on the metal surface of a photo-electric cell and a photo-electrci current is produced. The lens forming the image is then replaced by another of the same diameter but of focal length 15 cm. The photo-electric current in this case is

A
I/2

B
I

C
2I

D
4I

Solution

Lenses of the same diameter collect equal amounts of light.
Intensity is the measure of amount of light collected, hence the intensity remains the same.
Intensity is measured by the photoelectric current. So the photoelectric current would also remain the same.
Question 5
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A proton and an electron are accelerated by same potential difference have de-Broglie wavelength l_p and l_e .

A
l_e = l_p

B
l_e p

C
l_e >l_p

D
none of these

Solution

Charge present on both particles are same
∵λ= h /mqV so q₁ =q₂
λ_e /λ_p = √(m_p v /m_e v) =>1
∵m_p >>m_e
∴λ_e >λ_p .
Option C is the correct answer.
Question 6
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An electron with initial kinetic energy of 100eV is acceleration through a potential difference of 50V. Now the de-Broglie wavelength of electron becomes.

A
1 Å

B

C

D
12.27 Å

Solution
Question 7
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If h is Planck 's constant is SI system, the momentum of a photon of wavelength 0.01 Åis

A
10^-2 h

B
h

C
10² h

D
10¹² h

Solution

Momentum of photo, p=E/c =h ν/c where E is the energy of a photon and c is the velocity of light.
∴p= hc/c λ [∵ν=λc ]
p=h/λ=h/(0.01 ×10 −10)=10¹² h
Question 8
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The angular momentum of an electron in the hydrogen atom is . Here h is Planck 's constant. The kinetic energy of this electron is

A
4.53 eV

B
1.51 eV

C
3.4 eV

D
6.8 eV

Solution

Angular momentum,
L=nh/2 π=3h /2 π(given)
also n=3
∴ E=(−1.36 / n² )eV
E=−13.6/(3)² eV
E=−13.6/9 eV
E=−1.57eV
Question 9
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Consider the following electronic energy level diagram of H-atom : Photons associated with shortest and longest wavelengths would be emitted from the atom by the transitions labelled.

A
D and C respectively

B
C and A respectively

C
C and D respectively

D
A and C respectively

Solution

ΔE=k[(1/n₂² )-1/n₁² ]=hc/λ
If transition from n-4->n=2
ΔE-->Maximum energy.
Wavelength λwill be the shortest for C.
If transition occurs from n=4--->n=3
Then, ΔE--->Minimum
Wavelength λwill be longest for D.
Question 10
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If the electron in a hydrogen atom were in the energy level with n = 3, how much energy in joule would be required to ionise the atom ? (Ionisation energy of H-atom is 2.18 ×10^-18 J)

A
6.54 ×10^-19

B
1.43 ×10^-19

C
2.42 ×10^-19

D
3.14 ×10^-20

Solution

Energy of e −in n^th shell =−13.6 (τ² /n² )
For H −atom with n=3 →E=−13.6 ×(1/9)
⇒E=1/9 (Energy of e −in first shell)
⇒E=(1/9)×2.18 ×10⁻¹⁸ )
⇒E=2.42 ×10⁻¹⁹ J(Ionisation Energy= Energy of e⁻ in first shell)