If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission?
| Metal | Work Function (eV) |
| Na | 1.92 |
| K | 2.15 |
| Ca | 3.20 |
| Mo | 4.17 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Na and K only
Photoelectric emission occurs only if the photon energy is greater than the work function of the metal. Find the photon energy, using \( hc = 1240\,\mathrm{eV\,nm} \):
\( \displaystyle E = \frac{hc}{\lambda} = \frac{1240\,\mathrm{eV\,nm}}{412.5\,\mathrm{nm}} \approx 3.01\,\mathrm{eV} \)
Compare with each work function:
\( \text{Na: } 1.92\,\mathrm{eV} \lt 3.01\,\mathrm{eV} \;\text{(emits)} \)
\( \text{K: } 2.15\,\mathrm{eV} \lt 3.01\,\mathrm{eV} \;\text{(emits)} \)
\( \text{Ca: } 3.20\,\mathrm{eV} \gt 3.01\,\mathrm{eV} \;\text{(no emission)} \)
\( \text{Mo: } 4.17\,\mathrm{eV} \gt 3.01\,\mathrm{eV} \;\text{(no emission)} \)
So only Na and K show photoelectric emission.