Question

Radiation of wavelength $331\text{ nm}$ irradiates the following metals:

• A. Only Na and K show photoelectric emission.
• B. Only Mo will not show photoelectric emission.
• C. All of the given metals show photoelectric emission.
• D. None of them show photoelectric emission.

Hint:

Use the shortcut formula $E\text{ (in eV)} \approx \frac{1240}{\lambda\text{ (in nm)}}$ for rapid estimation during competitive exams. Here, $E \approx \frac{1240}{331} \approx 3.75\text{ eV}$.

Answer 1

The Correct Option is B

Concept:
• Photoelectric emission occurs from a metal surface if and only if the energy of incident photon $E$ is greater than or equal to the work function $\Phi_0$ of that metal.

• The energy of an incident photon of wavelength $\lambda$ is given by $E = \frac{hc}{\lambda}$.

Step 1: Calculate the energy of incident photon
Given wavelength of incident radiation $\lambda = 331\text{ nm} = 331 \times 10^{-9}\text{ m}$.
Substitute Planck's constant $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$ and speed of light $c = 3 \times 10^8\text{ m/s}$:
\[ E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9}}\text{ J} \]
Convert energy from Joules to electron-volts ($1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$):
\[ E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9} \times 1.6 \times 10^{-19}}\text{ eV} \]
\[ E = \frac{1.989 \times 10^{-25}}{5.296 \times 10^{-26}}\text{ eV} \approx 3.755\text{ eV} \]

Step 2: Compare photon energy with work functions
The incident photon energy is $E \approx 3.75\text{ eV}$.
Comparing $E$ with the given work functions ($\Phi_0$):
For Sodium (Na): $\Phi_0 = 1.92\text{ eV} < 3.75\text{ eV}$ $\implies$ Photoelectric emission occurs.
For Potassium (K): $\Phi_0 = 2.15\text{ eV} < 3.75\text{ eV}$ $\implies$ Photoelectric emission occurs.
For Calcium (Ca): $\Phi_0 = 3.20\text{ eV} < 3.75\text{ eV}$ $\implies$ Photoelectric emission occurs.
For Molybdenum (Mo): $\Phi_0 = 4.17\text{ eV} > 3.75\text{ eV}$ $\implies$ Photoelectric emission DOES NOT occur.

Step 3: Conclusion
Among the four metals, only Molybdenum (Mo) has a work function higher than the incident photon energy. Therefore, only Mo will not show photoelectric emission, corresponding to option (B).