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Origin Of Quantum Mechanics NAT Level – 2

10 Questions MCQ Test Modern Physics for IIT JAM | Origin Of Quantum Mechanics NAT Level – 2

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Attempt Origin Of Quantum Mechanics NAT Level – 2 | 10 questions in 45 minutes | Mock test for IIT JAM preparation | Free important questions MCQ to study Modern Physics for IIT JAM for IIT JAM Exam | Download free PDF with solutions

*Answer can only contain numeric values

QUESTION: 1

The life time of a nucleus in excited state is 10^–12s. The uncertainty in the energy and frequency of γ-ray photon emitted by it. (h = 6.63 × 10^–34 Js) (in units of 10¹¹ Hz) is a × 10¹¹Hz. Find a.

Solution:

From the energy time uncertainty

The correct answer is: 1.59 x 10^11 Hz

*Answer can only contain numeric values

QUESTION: 2

The number of photon emitted per second by a 40W source of mono chromatic light of wavelength 6000Å is given by x × 10²⁰. Find the value of x.

Solution:

The correct answer is: 1.206

*Answer can only contain numeric values

QUESTION: 3

For what wavelength of photon does compton scattering result in a photon whose energy is one half that of the original photon at a scattering angle of 45º? Give the answer as [a × 10^–3 Å] . Find the value of a.

Solution:

The correct answer is: 7.085

*Answer can only contain numeric values

QUESTION: 4

For what value of an electron’s speed will its de-Broglie wavelength be same as the compton wavelength? (in terms of c)

Solution:

de Broglie wavelength

Compton wavelength

The correct answer is: 0.707

*Answer can only contain numeric values

QUESTION: 5

Gamma ray photons of energy 1.02 MeV are scattered from electrons which are initially at rest. Find the angle for symmetric scattering at this energy (in degrees)

Solution:

For symmetric scattering θ = φ
From the relation between θ & φ

The correct answer is: 41.4

*Answer can only contain numeric values

QUESTION: 6

The average life time of an excited atomic state is 10^–8s. If the wavelength of the spectral line associated with the transition from this state to the ground state is 6000 Å. What will be the width of this line? (in femtometer)

Solution:

Average life time = Δt = uncertainty in time = 10^–8s

width of the line = 1.9 × 10^–14 m
The correct answer is: 19

*Answer can only contain numeric values

QUESTION: 7

If the X-ray photon is scattered at angle of 180° and electron recoils with an energy of 4 keV. Then calculate the wavelength of the incident photon (in Å) in angstroms.

Solution:

K.E. of recoiled e^–is
E = 4 × 10³ × 1.6 × 10^–19 Joules = 6.4 × 10^–16 J

= 34.13 × 10^–24 kg m/s

From conservation of momentum

solving (i) & (ii)

The correct answer is: 0.35

*Answer can only contain numeric values

QUESTION: 8

The minimum kinetic energy of an alpha particle that can exist in a nucleus (use uncertainty principle ) (Radius of nucleus = 10^–14m) (in keV)

Solution:

Radius of nucleus = 10^–14 m
∴ Uncertainty in position

= 1.054 × 10^–20 kg m/s
Mass of α particle = 4m_p = 4 × 1.67 × 10^–27 kg
0
The correct answer is: 52

*Answer can only contain numeric values

QUESTION: 9

The intensity of scattered monochromatic beam of X-rays is plotted as a function of wavelength. There are 2 peaks observed. The distance between the 2 peaks is 0.024 Å. Find the angle (in degrees) at which the X-rays are scattered.

Solution:

Distance between two peaks = Δλ

The correct answer is: 90

*Answer can only contain numeric values

QUESTION: 10

Calculate the ground state energy of a Helium atom, using the uncertainty principle. (in eV)

Solution:

For a helium atom

Because momentum of electron 1,
and momentum of electron 2,
p₁, p₂ is the spread in momentum corresponding to electron 1 & 2 respectively.

r₁, r₂ is the localization of electron 1 & 2 respectively.

Total energy, E = K.E + Interaction energy of e₁ and e₂ + Interaction energy between nucleus and electrons

Interaction Energy between
Since the separation between r₁ and r₂ is of the order r₁ + r₂ and k = 1 in natural system of units.
Interaction between nucleus and electron

For the ground state energy
since in the ground state, Energy is minimum.

solving these two equations, we get

substituting r₁ and r₂ in the expression for E

E_min = –10.34 eV
Put m = 9.1 × 10^–31 kg
e = 1.6 × 10^–19C

The correct answer is: -10.34