Ultraviolet light of wavelength $2271 \,\mathring{A}$ from a $100 \; W$ mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is $-1.3 \; V$, estimate the work function of the metal. How would the photo-cell respond to a high-intensity $(10^{5} \; W \; m^{-2})$ red light of wavelength $6328 \,\mathring{A}$ produced by a $He-Ne$ laser?

(N/A) Given:
Wavelength of ultraviolet light, $\lambda = 2271 \,\mathring{A} = 2271 \times 10^{-10} \, m$
Stopping potential, $V_{0} = 1.3 \, V$
Planck's constant, $h = 6.63 \times 10^{-34} \, J \cdot s$
Speed of light, $c = 3 \times 10^{8} \, m/s$
Charge of an electron, $e = 1.6 \times 10^{-19} \, C$
Using Einstein's photoelectric equation: $\phi_{0} = \frac{hc}{\lambda} - eV_{0}$
$\phi_{0} = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^{8})}{2271 \times 10^{-10}} - (1.6 \times 10^{-19} \times 1.3)$
$\phi_{0} = 8.758 \times 10^{-19} \, J - 2.08 \times 10^{-19} \, J = 6.678 \times 10^{-19} \, J$
In electron volts: $\phi_{0} = \frac{6.678 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 4.17 \, eV$
For red light: $\lambda_{r} = 6328 \,\mathring{A} = 6328 \times 10^{-10} \, m$
Energy of red light photon, $E_{r} = \frac{hc}{\lambda_{r}} = \frac{1.989 \times 10^{-25}}{6328 \times 10^{-10}} \approx 3.14 \times 10^{-19} \, J \approx 1.96 \, eV$
Since $E_{r} < \phi_{0}$ $(1.96 \, eV < 4.17 \, eV)$, the red light will not cause photoelectric emission, regardless of its high intensity.