Modern Physics - NEET Physics Chapterwise MCQs & PYQs

NEET Modern Physics MCQs & PYQs

Practice NEET Modern Physics Questions

Question 61:

easy

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): Nucleus having more binding energy per nucleon is more stable.

Reason (R): Stability increase with increase in number of nucleons.

In the light of the above statements, choose the correct answer from the options given below.

Assertion is true because binding energy per nucleon is the direct measure of nuclear stability. Reason is false because stability does not simply increase with nucleon number; heavy nuclei with a very large number of nucleons become unstable.

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Question 62:

easy

A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is

Initial BE \(= 240 \times 7.6 = 1824\) MeV. Final BE \(= 2 \times (120 \times 8.5) = 2040\) MeV. Total gain in BE \(= 2040 - 1824 = 216\) MeV.

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Question 63:

easy

The number of photons per second on an average emitted by the source of monochromatic light of wavelength \(600\text{ nm}\), when it delivers the power of \(3.3 \times 10^{-3}\text{ watt}\) will be (\(h = 6.6 \times 10^{-34}\text{ J s}\))

Power \(P = n \frac{hc}{\lambda}\), where \(n\) is the number of photons emitted per second. Substituting the given values, we get \(n = \frac{3.3 \times 10^{-3} \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 10^{16}\text{ s}^{-1}\).

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Question 64:

easy

An electromagnetic wave of wavelength \(lambda\) is incident on a photosensitive surface of negligible work function. If \(m\) is mass of photoelectron emitted from the surface has de-Broglie wavelength \(lambda_d\), then

With a negligible work function, the maximum kinetic energy of the emitted photoelectron is \(E = \frac{hc}{\lambda}\) and its de-Broglie wavelength is \(\lambda_d = \frac{h}{\sqrt{2mE}}\). Substituting \(E\) gives \(\lambda_d^2 = \frac{h\lambda}{2mc}\), which simplifies to \(\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2\).

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Question 65:

easy

The potential difference that must be applied to stop the fastest moving photoelectrons emitted by a metal surface, having work function \(4.5\text{ eV}\), when ultraviolet light of \(2000A^\circ\) falls on it, will be

Formula: \(eV_0 = E - \Phi_0\), where \(E = \frac{hc}{\lambda} = \frac{12400}{2000} = 6.2\text{ eV}\). Thus, the stopping potential \(V_0 = 6.2 - 4.5 = 1.7\text{ V}\), requiring an applied potential of \(-1.7\text{ V}\).

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Question 66:

easy

An electron in the hydrogen atom jumps from state \(n\) to the ground state. The wavelength so emitted illuminates a photo-sensitive material having work function \(4.09\text{ eV}\). If the stopping potential of the photoelectrons is \(8\text{ V}\), the value of \(n\) is

Formula: \(E = K_{max} + \Phi_0 = 8 + 4.09 = 12.09\text{ eV}\). For hydrogen atom transition, \(13.6 \left(1 - \frac{1}{n^2}\right) = 12.09\). Solving this gives \(n = 3\).

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Question 67:

easy

When \(^{235}_{92}\text{U}\) undergoes fission, 0.2% of its original mass is changed into energy. How much energy is released if 1 kg of \(^{235}_{92}\text{U}\) undergoes fission?

Formula: \(E = \Delta m c^2\). The mass defect is \(\Delta m = 0.2\%\ of 1 kg = 0.002\text{ kg}\). Thus, \(E = 0.002 \times (3 \times 10^8)^2 = 18 \times 10^{13}\text{ J}\).

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Question 68:

easy

The mass of a proton is \(1.0073\text{ u}\) and that of neutron is \(1.0087\text{ u}\) (\(u = \text{atomic mass unit}\)). The binding energy of \({}_2\text{He}^4\) is (mass of helium nucleus = \(4.0015\text{ u}\))

Mass defect \(\Delta m = (2 m_p + 2 m_n) - m_{\text{He}} = [2(1.0073) + 2(1.0087)] - 4.0015 = 0.0305\text{ u}\). Binding energy is \(E_b = \Delta m \times 931.5\text{ MeV} \approx 0.0305 \times 931.5 \approx 28.4\text{ MeV}\).

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Question 69:

easy

Consider a hypothetical hydrogen like atom. Wavelength of spectral lines corresponding to transition of electron from \( n = m \) to \( n = 1 \) are given by \( \lambda = \frac{1800 m^2}{m^2 – 1} \text{ \A^\circ} \) \( (m = 2, 3, 4, 5, \dots, \infty) \). Minimum energy of emitted photon is nearly

Minimum energy corresponds to the maximum wavelength, which occurs at \( m = 2 \). Thus, \( \lambda_{\max} = \frac{1800 \times 2^2}{2^2 - 1} = 2400 \text{ \A^\circ} \). The minimum energy is \( E_{\min} = \frac{12400}{\lambda_{\max}} = \frac{12400}{2400} \approx 5.17 \text{ eV} \approx 5.2 \text{ eV} \).

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Question 70:

easy

Beta positive decay is

In \( \beta^+ \) decay, a proton inside the nucleus converts into a neutron, emitting a positron (\( e^+ \)) and a neutrino (\( \nu_e \)). The nuclear reaction is \( p \rightarrow n + e^+ + \nu_e \).

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