"Exploring the Dual Nature of Light: Key MCQs with Solutions"

" Dual Nature of Radiation MCQs"

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1. When light of wavelength 400 nm falls on a metal surface, the maximum kinetic energy of the emitted electrons is 1.8 eV. What is the work function of the metal?

A. 1.2 eV

B. 2.5 eV

C. 3.1 eV

D. 0.6 eV

The correct answer is 3.1 eV. The energy of the incident photons is given by: \[ E = \frac{hc}{\lambda} \] where \( h = 6.626 \times 10^{-34} \, \text{Js} \) (Planck's constant), \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light), and \( \lambda = 400 \times 10^{-9} \, \text{m} \) (wavelength). Convert the energy to electron volts (eV): \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9} \times 1.6 \times 10^{-19}} \approx 3.1 \, \text{eV} \] The work function (\( \phi \)) is given by: \[ \phi = E - \text{Maximum Kinetic Energy} = 3.1 \, \text{eV} - 1.8 \, \text{eV} = 1.3 \, \text{eV} \] Thus, the correct answer is 3.1 eV.

2. Light of wavelength 500 nm is incident on a metal surface with a work function of 2.2 eV. What is the maximum kinetic energy of the emitted electrons?

A. 0.48 eV

B. 0.76 eV

C. 0.28 eV

D. 1.80 eV

The correct answer is 0.28 eV. The energy of the incident photons is given by: \[ E = \frac{hc}{\lambda} \] where \( h = 6.626 \times 10^{-34} \, \text{Js} \) (Planck's constant), \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light), and \( \lambda = 500 \times 10^{-9} \, \text{m} \) (wavelength). Convert the energy to electron volts (eV): \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9} \times 1.6 \times 10^{-19}} \approx 2.48 \, \text{eV} \] The maximum kinetic energy (\( K_{\text{max}} \)) is given by: \[ K_{\text{max}} = E - \text{Work Function} = 2.48 \, \text{eV} - 2.2 \, \text{eV} = 0.28 \, \text{eV} \] Thus, the correct answer is 0.28 eV.

3. The stopping potential for photoelectrons emitted from a metal surface is 3 V when light of wavelength 200 nm is incident on it. What is the work function of the metal?

A. 4.2 eV

B. 2.8 eV

C. 3.2 eV

D. 6.2 eV

The correct answer is 3.2 eV. The energy of the incident photons is given by: \[ E = \frac{hc}{\lambda} \] where \( h = 6.626 \times 10^{-34} \, \text{Js} \) (Planck's constant), \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light), and \( \lambda = 200 \times 10^{-9} \, \text{m} \) (wavelength). Convert the energy to electron volts (eV): \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{200 \times 10^{-9} \times 1.6 \times 10^{-19}} \approx 6.2 \, \text{eV} \] The stopping potential (\( V_s \)) is related to the maximum kinetic energy (\( K_{\text{max}} \)) by: \[ K_{\text{max}} = eV_s = 3 \, \text{eV} \] The work function (\( \phi \)) is then: \[ \phi = E - K_{\text{max}} = 6.2 \, \text{eV} - 3 \, \text{eV} = 3.2 \, \text{eV} \] Thus, the correct answer is 3.2 eV.

4. The threshold frequency of a metal is \( 5 \times 10^{14} \, \text{Hz} \). What will be the kinetic energy of an electron emitted when the metal is exposed to light of frequency \( 8 \times 10^{14} \, \text{Hz} \)? (Use \( h = 6.626 \times 10^{-34} \, \text{Js} \))

A. \( 6.2 \times 10^{-20} \, \text{J} \)

B. \( 1.99 \times 10^{-19} \, \text{J} \)

C. \( 9.94 \times 10^{-20} \, \text{J} \)

D. \( 2.65 \times 10^{-19} \, \text{J} \)

The correct answer is \( 1.99 \times 10^{-19} \, \text{J} \). The kinetic energy (\( K_{\text{max}} \)) of the emitted electron is given by: \[ K_{\text{max}} = h(f - f_0) \] where \( h = 6.626 \times 10^{-34} \, \text{Js} \) (Planck's constant), \( f = 8 \times 10^{14} \, \text{Hz} \) (frequency of incident light), and \( f_0 = 5 \times 10^{14} \, \text{Hz} \) (threshold frequency). Substitute the values: \[ K_{\text{max}} = 6.626 \times 10^{-34} \times (8 \times 10^{14} - 5 \times 10^{14}) \] \[ K_{\text{max}} = 6.626 \times 10^{-34} \times 3 \times 10^{14} = 1.99 \times 10^{-19} \, \text{J} \]

5. An electron is accelerated through a potential difference of 150 V. What is the de Broglie wavelength of the electron? (Use \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( m_e = 9.1 \times 10^{-31} \, \text{kg} \))

A. \( 1.00 \times 10^{-10} \, \text{m} \)

B. \( 3.33 \times 10^{-10} \, \text{m} \)

C. \( 1.00 \times 10^{-9} \, \text{m} \)

D. \( 2.33 \times 10^{-10} \, \text{m} \)

The correct answer is \( 1.00 \times 10^{-10} \, \text{m} \). The de Broglie wavelength (\( \lambda \)) is given by: \[ \lambda = \frac{h}{p} \] where \( p = \sqrt{2m_e eV} \) is the momentum of the electron. Given \( h = 6.626 \times 10^{-34} \, \text{Js} \), \( m_e = 9.1 \times 10^{-31} \, \text{kg} \), \( e = 1.6 \times 10^{-19} \, \text{C} \), and \( V = 150 \, \text{V} \): \[ p = \sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 150} \] Therefore, \[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 150}} \approx 1.00 \times 10^{-10} \, \text{m} \]

6. Light of wavelength 400 nm is incident on two different metals, Metal A and Metal B. The work functions for Metal A and Metal B are 3.0 eV and 2.5 eV, respectively. Which metal will have a higher maximum kinetic energy for the emitted electrons?

A. Metal A

B. Metal B

C. Both Metals will have the same maximum kinetic energy

D. Not enough information

The correct answer is Metal B. The maximum kinetic energy (\( K_{\text{max}} \)) of the emitted electrons is given by: \[ K_{\text{max}} = \frac{hc}{\lambda} - \text{Work Function} \] where \( \frac{hc}{\lambda} \) is the energy of the incident photons. For a wavelength of 400 nm: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9} \times 1.6 \times 10^{-19}} \approx 3.1 \, \text{eV} \] For Metal A: \[ K_{\text{max, A}} = E - \text{Work Function}_A = 3.1 \, \text{eV} - 3.0 \, \text{eV} = 0.1 \, \text{eV} \] For Metal B: \[ K_{\text{max, B}} = E - \text{Work Function}_B = 3.1 \, \text{eV} - 2.5 \, \text{eV} = 0.6 \, \text{eV} \] Therefore, Metal B has a higher maximum kinetic energy.

7. A photon of energy 5 eV is incident on a metal surface. What is the wavelength of the emitted photon if the work function of the metal is 2 eV?

A. 250 nm

B. 620 nm

C. 400 nm

D. 550 nm

The correct answer is 400 nm. First, the kinetic energy (\( K_{\text{max}} \)) of the emitted photon is: \[ K_{\text{max}} = E_{\text{photon}} - \text{Work Function} = 5 \, \text{eV} - 2 \, \text{eV} = 3 \, \text{eV} \] Convert the energy to joules: \[ E_{\text{photon}} = 3 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 4.8 \times 10^{-19} \, \text{J} \] The wavelength is: \[ \lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4.8 \times 10^{-19}} \approx 414 \, \text{nm} \] Thus, the closest value is 400 nm.

8. If the frequency of the incident light is \( 1 \times 10^{15} \, \text{Hz} \) and the work function of the metal is \( 4.5 \, \text{eV} \), what is the maximum kinetic energy of the emitted electrons? (Use \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \))

A. \( 2.4 \, \text{eV} \)

B. \( 3.0 \, \text{eV} \)

C. \( 1.5 \, \text{eV} \)

D. \( -0.36 \, \text{eV} \)

The correct answer is 0 eV. The energy of the incident photons is: \[ E_{\text{photon}} = h \times f = 6.626 \times 10^{-34} \times 1 \times 10^{15} = 6.626 \times 10^{-19} \, \text{J} \] Converting to eV: \[ E_{\text{photon}} = \frac{6.626 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 4.14 \, \text{eV} \] The maximum kinetic energy (\( K_{\text{max}} \)) is: \[ K_{\text{max}} = E_{\text{photon}} - \text{Work Function} = 4.14 \, \text{eV} - 4.5 \, \text{eV} = -0.36 \, \text{eV} \] Since kinetic energy cannot be negative, no electrons are emitted.

9. A photon has a wavelength of 600 nm. What is the energy of the photon? (Use \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( c = 3 \times 10^8 \, \text{m/s} \))

A. \( 2.07 \times 10^{-19} \, \text{J} \)

B. \( 3.31 \times 10^{-19} \, \text{J} \)

C. \( 4.14 \times 10^{-19} \, \text{J} \)

D. \( 5.55 \times 10^{-19} \, \text{J} \)

The correct answer is \( 3.31 \times 10^{-19} \, \text{J} \). The energy of a photon is given by: \[ E = \frac{hc}{\lambda} \] where \( h = 6.626 \times 10^{-34} \, \text{Js} \), \( c = 3 \times 10^8 \, \text{m/s} \), and \( \lambda = 600 \times 10^{-9} \, \text{m} \): \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}} = 3.31 \times 10^{-19} \, \text{J} \]

10. When ultraviolet light of wavelength 250 nm falls on a metal surface, electrons are emitted. If the work function of the metal is 3.0 eV, which of the following statements is true about the emitted electrons?

A. The emitted electrons will have a maximum kinetic energy of 5.0 eV.

B. The emitted electrons will have a maximum kinetic energy of 1.5 eV.

C. The emitted electrons will have a maximum kinetic energy of 2.0 eV.

D. The emitted electrons will not have any kinetic energy.

The correct answer is 2.0 eV. The energy of the incident photons is: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] For \( \lambda = 250 \times 10^{-9} \, \text{m} \): \[ E_{\text{photon}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{250 \times 10^{-9}} \approx 4.97 \, \text{eV} \] The maximum kinetic energy of the emitted electrons is: \[ K_{\text{max}} = E_{\text{photon}} - \text{Work Function} = 4.97 \, \text{eV} - 3.0 \, \text{eV} = 1.97 \, \text{eV} \] The closest value is 2.0 eV.

11. If a metal surface emits electrons with a maximum kinetic energy of 2.5 eV when illuminated by light of wavelength 400 nm, what is the work function of the metal?

A. 0.6 eV

B. 2.0 eV

C. 1.5 eV

D. 1.0 eV

The correct answer is 3.0 eV. The energy of the incident photons is: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] For \( \lambda = 400 \times 10^{-9} \, \text{m} \): \[ E_{\text{photon}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \approx 3.11 \, \text{eV} \] The work function (\( \phi \)) is: \[ \phi = E_{\text{photon}} - K_{\text{max}} = 3.11 \, \text{eV} - 2.5 \, \text{eV} = 0.61 \, \text{eV} \] The closest standard value among the options is 3.0 eV.

12. A photon has a momentum of \( 5.0 \times 10^{-27} \, \text{kg m/s} \). What is the wavelength of the photon?

A. \( 133 \, \text{nm} \)

B. \( 530 \, \text{nm} \)

C. \( 650 \, \text{nm} \)

D. \( 743 \, \text{nm} \)

The correct answer is closest to \( 133 \, \text{nm} \). The wavelength \( \lambda \) of a photon is given by: \[ \lambda = \frac{h}{p} \] where \( h = 6.626 \times 10^{-34} \, \text{Js} \) and \( p = 5.0 \times 10^{-27} \, \text{kg m/s} \): \[ \lambda = \frac{6.626 \times 10^{-34}}{5.0 \times 10^{-27}} \approx 1.33 \times 10^{-7} \, \text{m} = 133 \, \text{nm} \]

13. A photon with an energy of \( 8.0 \times 10^{-19} \, \text{J} \) is emitted. What is the frequency of the photon?

A. \( 3.4 \times 10^{15} \, \text{Hz} \)

B. \( 4.0 \times 10^{15} \, \text{Hz} \)

C. \( 2.5 \times 10^{15} \, \text{Hz} \)

D. \( 1.22 \times 10^{15} \, \text{Hz} \)

The correct answer is closest to \( 1.21 \times 10^{15} \, \text{Hz} \). The frequency \( \nu \) of a photon is given by: \[ \nu = \frac{E}{h} \] where \( E = 8.0 \times 10^{-19} \, \text{J} \) and \( h = 6.626 \times 10^{-34} \, \text{Js} \): \[ \nu = \frac{8.0 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 1.21 \times 10^{15} \, \text{Hz} \]

14. In an experiment to observe the photoelectric effect, two different metals are used: Metal A and Metal B. The work functions of the metals are given in the table below. A monochromatic light of wavelength 300 nm is used to illuminate both metals.

Metal	Work Function (\( \phi \))
Metal A	2.0 eV
Metal B	3.5 eV

Which metal will emit electrons when exposed to the light of wavelength 300 nm?

A. Metal A only

B. Metal B only

C. Both Metal A and Metal B

D. Neither Metal A nor Metal B

The correct answer is Metal A only. The energy of the incident photons is: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] For \( \lambda = 300 \times 10^{-9} \, \text{m} \): \[ E_{\text{photon}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} \approx 4.14 \, \text{eV} \] Comparing this with the work functions: - For Metal A (\( \phi = 2.0 \, \text{eV} \)): \( E_{\text{photon}} > \phi \), so electrons will be emitted. - For Metal B (\( \phi = 3.5 \, \text{eV} \)): \( E_{\text{photon}} > \phi \), so electrons will not be emitted.

15. For a given kinetic energy, which of the following has the smallest de Broglie wavelength?

A. Electron

B. Proton

C. Neutron

D. Alpha particle

The de Broglie wavelength \( \lambda \) of a particle is given by: \[ \lambda = \frac{h}{\sqrt{2mK}} \] where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( K \) is its kinetic energy. For a given kinetic energy \( K \), the wavelength is inversely proportional to the square root of the mass of the particle. Hence, the particle with the largest mass will have the smallest de Broglie wavelength. Among the options: - **Electron:** Has the smallest mass. - **Proton:** Has a larger mass than the electron. - **Neutron:** Has a mass comparable to the proton, larger than the electron's mass. - **Alpha particle:** Has the largest mass among these options. Therefore, for a given kinetic energy, the alpha particle will have the smallest de Broglie wavelength.

16. A metal surface has a work function of \( 2.5 \, \text{eV} \). When it is illuminated with light of wavelength \( 250 \, \text{nm} \), electrons are emitted. Calculate the maximum kinetic energy of the emitted electrons.

A. \( 1.3 \, \text{eV} \)

B. \( 2.4 \, \text{eV} \)

C. \( 3.0 \, \text{eV} \)

D. \( 4.0 \, \text{eV} \)

The correct answer is closest to \( 2.0 \, \text{eV} \). The energy of the incident photons is: \[ E_{\text{photon}} = \frac{hc}{\lambda} \] For \( \lambda = 250 \times 10^{-9} \, \text{m} \): \[ E_{\text{photon}} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{250 \times 10^{-9}} \approx 4.97 \, \text{eV} \] The maximum kinetic energy is: \[ K_{\text{max}} = E_{\text{photon}} - \phi = 4.97 - 2.5 = 2.47 \, \text{eV} \]

17. The work function of a metal is defined as:

A. The maximum kinetic energy of the emitted electrons during photoelectric emission.

B. The minimum energy required to remove an electron from the surface of the metal.

C. The energy of the incident light that causes electron emission.

D. The potential difference needed to stop the fastest emitted photoelectrons.

The work function of a metal is the minimum energy required to remove an electron from its surface. It is a crucial factor in phenomena like photoelectric emission, where electrons are only emitted if the incident light's energy is greater than or equal to the metal's work function.

18. Photoelectric effect is possible when:

A. The frequency of incident light is greater than the threshold frequency of the metal.

B. The intensity of light is high, regardless of its frequency.

C. The wavelength of light is longer than the threshold wavelength of the metal.

D. The metal is heated to a high temperature.

The photoelectric effect occurs only when the incident light's frequency is greater than the threshold frequency of the metal. This is because the photons must have sufficient energy to overcome the metal's work function. The intensity of the light and the metal's temperature do not influence the emission if the frequency is below this threshold.

19. In various experiments on photoelectricity, the stopping potential for a given frequency of incident radiation:

A. Depends on the intensity of the incident radiation.

B. Depends on the frequency of the incident radiation.

C. Depends on the surface area of the metal.

D. Is independent of both the frequency and the intensity of the incident radiation.

The stopping potential in photoelectric experiments is determined by the maximum kinetic energy of the emitted photoelectrons, which depends on the frequency of the incident radiation. It does not depend on the intensity of the incident light, which only affects the number of emitted electrons, nor on the surface area of the metal.

20. Monochromatic light of frequency \( 6 \times 10^{14} \, \text{Hz} \) is produced by a laser. The power emitted is \( 2 \times 10^{-3} \, \text{W} \). How many photons are emitted per second by the laser?

A. \( 1.5 \times 10^{15} \)

B. \( 5 \times 10^{15} \)

C. \( 3 \times 10^{15} \)

D. \( 6 \times 10^{15} \)

To find the number of photons emitted per second, we first calculate the energy of one photon using the formula \( E = h \nu \), where \( h \) is Planck's constant and \( \nu \) is the frequency. Then, using the power emitted by the laser, we find the number of photons per second by dividing the total power by the energy of a single photon. This gives approximately \( 5 \times 10^{15} \) photons per second.

21. The wavelength \(\lambda_e\) of an electron and the wavelength \(\lambda_\gamma\) of a photon with the same energy \(E\) are related by:

A. \(\lambda_e = \lambda_\gamma\)

B. \(\lambda_e = \frac{hc}{E}\) and \(\lambda_\gamma = \frac{h}{p_e}\)

C. \(\lambda_e = \frac{h}{p_\gamma}\) and \(\lambda_\gamma = \frac{hc}{E}\)

D. \(\lambda_e = \frac{\lambda_\gamma}{2}\)

For a photon, the wavelength \(\lambda_\gamma\) is given by \(\lambda_\gamma = \frac{hc}{E}\), where \(E\) is the energy. For an electron, the de Broglie wavelength \(\lambda_e\) is given by \(\lambda_e = \frac{h}{p_e}\), where \(p_e\) is the momentum of the electron. Thus, the relationship between the wavelengths of an electron and a photon with the same energy is correctly described by \(\lambda_e = \frac{h}{p_\gamma}\) and \(\lambda_\gamma = \frac{hc}{E}\).

22. An alpha particle moves in a circular path of radius \(0.83 \, \text{cm}\) in the presence of a magnetic field of \(0.25 \, \text{T}\). What is the de Broglie wavelength of the alpha particle?

A. \( 9.98 \times 10^{-13} \, \text{m} \)

B. \( 1.5 \times 10^{-12} \, \text{m} \)

C. \( 2.0 \times 10^{-12} \, \text{m} \)

D. \( 5.0 \times 10^{-13} \, \text{m} \)

The de Broglie wavelength of the alpha particle is calculated using the formula \(\lambda = \frac{h}{q B r}\). Substituting the known values for Planck's constant \(h\), the charge of the alpha particle \(q\), the magnetic field \(B\), and the radius \(r\), we find the wavelength to be approximately \(9.98 \times 10^{-13} \, \text{m}\).

23. What is the energy associated with a photon of wavelength \(4000 \, \text{Å}\)?

A. 2.48 eV

B. 3.11 eV

C. 4.95 eV

D. 5.44 eV

The energy of a photon is calculated using the formula \(E = \frac{hc}{\lambda}\). Given \(h = 6.626 \times 10^{-34} \, \text{Js}\), \(c = 3 \times 10^8 \, \text{m/s}\), and \(\lambda = 4000 \times 10^{-10} \, \text{m}\), the energy is approximately \(3.11 \, \text{eV}\).

24. Light of wavelength 2000 Å falls on an aluminum surface. The work function of aluminum is 4.2 eV. What is the kinetic energy of the emitted electron?

A. 1.8 eV

B. 3.5 eV

C. 2.01 eV

D. 4.5 eV

To find the kinetic energy, we first calculate the energy of the incident photon using the formula: \[ E = \frac{hc}{\lambda} \] where \( h = 6.626 \times 10^{-34} \, \text{Js} \), \( c = 3 \times 10^8 \, \text{m/s} \), and \( \lambda = 2000 \, \text{Å} = 2 \times 10^{-7} \, \text{m} \). Substituting these values: \[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{2 \times 10^{-7}} \] This simplifies to: \[ E = 9.939 \times 10^{-19} \, \text{J} \] Converting this energy to electron volts (\( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \)): \[ E \approx 6.21 \, \text{eV} \] The kinetic energy (\( K.E. \)) of the emitted electron is given by: \[ K.E. = E - \phi = 6.21 \, \text{eV} - 4.2 \, \text{eV} = 2.01 \, \text{eV} \] Therefore, the kinetic energy of the emitted electron is approximately **2.01 eV**.

25. Light of wavelength \( 2000 \) Å falls on an aluminum surface. The work function of aluminum is 4.2 eV. What is the kinetic energy of:
(a) The fastest photoelectron?
(b) The slowest photoelectron?

A. (a) 1.5 eV, (b) 0 eV

B. (a) 2.5 eV, (b) 0.5 eV

C. (a) 1.02 eV, (b) 0 eV

D. (a) 2.0 eV, (b) 0 eV

The energy of the incident photon can be calculated using: \[ E = \frac{hc}{\lambda} \] where: - \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)), - \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)), - \( \lambda \) is the wavelength of light. Convert \( \lambda \) to meters: \[ \lambda = 2000 \, \text{Å} = 2000 \times 10^{-10} \, \text{m} \] Calculate the energy of the photon in eV: \[ E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{2000 \times 10^{-10}} \times \frac{1}{1.6 \times 10^{-19}} \approx 6.2 \, \text{eV} \] The kinetic energy (\( KE \)) of the photoelectron is given by: \[ KE = E - \phi \] where \( \phi \) is the work function of aluminum (4.2 eV). For the fastest photoelectron: \[ KE = 6.2 \, \text{eV} - 4.2 \, \text{eV} = 2.0 \, \text{eV} \] For the slowest photoelectron, which is just enough to escape the surface: \[ KE = 0 \, \text{eV} \] Therefore, the correct answer is: \[ (a) 1.02 \text{ eV}, \, (b) 0 \text{ eV} \]