Modern Physics

Practice Questions:

Question 21: 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 (\(h = 6.6 \times 10^{-34}\text{ J s}\))

1. \(10^{15}\)

2. \(10^{18}\)

3. \(10^{17}\)

4. \(10^{16}\)

View Answer

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 22: 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

1. \(\lambda = \left(\frac{2h}{mc}\right)\lambda_d^2\)

2. \(\lambda = \left(\frac{2m}{hc}\right)\lambda_d^2\)

3. \(\lambda_d = \left(\frac{2mc}{h}\right)\lambda^2\)

4. \(\lambda = \left(\frac{2mc}{h}\right)\lambda_d^2\)

View Answer

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 23: 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

1. -0.7 V

2. -1.7 V

3. -1.2 V

4. -0.8 V

View Answer

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 24: 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

1. 5

2. 2

3. 3

4. 4

View Answer

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 25: 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?

1. \[18 \times 10^{11} J\]

2. \[6 \times 10^{5} J\]

3. \[18 \times 10^{13} J\]

4. \[18 \times 10^{10} J\]

View Answer

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 26: 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 = \(4.0015\text{ u}\))

1. 28.4 MeV

2. 0.061 MeV

3. 3.05 MeV

4. 305 MeV

View Answer

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 27: 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

1. 5.2 eV

2. 7.03 eV

3. 10.2 eV

4. 13.6 eV

View Answer

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 28: easy

Beta positive decay is

1. The conversion of neutron into proton along with neutrino and electron

2. The conversion of proton into neutron along with antineutrino and electron

3. The conversion of neutron into proton along with positron and antineutrino

4. The conversion of proton into neutron along with neutrino and positron

View Answer

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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Question 29: easy

If \( E \) is the energy of \( n^{\text{th}} \) orbit of hydrogen atom, the energy of \( n^{\text{th}} \) orbit of \( \text{He}^+ \) ion will be

1. \( E \)

2. \( 2E \)

3. \( 3E \)

4. \( 4E \)

View Answer

The energy of an electron in a hydrogen-like atom is given by \( E_n \propto Z^2 \). For hydrogen, \( Z = 1 \), and for helium ion \( \text{He}^+ \), \( Z = 2 \). Therefore, \( E_{\text{He}^+} = Z^2 E = 4E \).

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Question 30: easy

In an experiment on photoelectric emission for incident light of wavelength \( 1.98 \times 10^{-7} \text{ m} \), stopping potential is found to be \( 2.5 \text{ V} \). What is maximum kinetic energy of emitted photoelectron?

1. 6.25 eV

2. 2.5 eV

3. 3.75 eV

4. Zero

View Answer

The maximum kinetic energy of emitted photoelectrons is related to the stopping potential by \( K_{\max} = e V_s \). Given \( V_s = 2.5 \text{ V} \), the maximum kinetic energy is simply \( 2.5 \text{ eV} \).

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