Ncert Exercises & Solutions

Question 11.26: Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photocell 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 high intensity ( $~$ 10⁵ W m^-2) red light of wavelength 6328 Å produced by a He-Ne laser?

Hint: Use Einstein's photoelectric equation.
Step1: Find the work function.
$\phi_0 = h \nu - eV_0\\=\frac{h c}{\lambda}-e V_{0}\\ =\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2271 \times 10^{-10}}-1.6 \times 10^{-19} \times 1.3\\\\=8.72 \times 10^{-19}-2.08 \times 10^{-19}\\ =6.64 \times 10^{-19} \mathrm{~J} \text {. }\\ =\frac{6.64 \times 10^{-19}}{1.6 \times 10^{-19}}\\ =4.15~eV.$
Step2: Find the threshold frequency of the metal.
$\begin{aligned} As,~~~~~~\phi_{0} &=h \nu_{0} \\ \Rightarrow \nu_{0} &=\frac{\phi_{0}}{h}=\frac{6.64 \times 10^{-19}}{6.6 \times 10^{-34}} \\ &=1.006 \times 10^{15} \mathrm{~Hz} . \end{aligned}$
Step 3: Find the frequency of red light.
$\begin{aligned} \nu_{r}=\frac{c}{_{\lambda_{r}}} &=\frac{3 \times 10^{8}}{6328 \times 10^{-10}} \\ &=4.74 \times 10^{14} \mathrm{~Hz} . \end{aligned}$
Since $\nu_0 > \nu , $ the photocell will not respond to the red light produced by the laser.

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