Ncert Exercises & Solutions

Question 11.11: Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.

Hint: Use Einstein’s photoelectric equation.
Step 1: Convert stopping potential in electron-volt.

Stopping the potential of the photoelectrons,
V₀=0.38 V.
We know that 1 eV = 1.6 x $10^{- 19}$ J.
$V_{0} = \frac{0.38}{1.6 \times 10^{- 19}} e V$
Step 2: Find the work function of the material.
$eV_{0}=\frac{hc}{\lambda}-\phi_{0}$

$\begin{array}{l} ϕ_{0} = \frac{h c}{λ} - e V_{0} \\ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{1.6 \times 10^{- 19} \times 488 \times 10^{- 9}} - \frac{1.6 \times 10^{- 19} \times 0.38}{1.6 \times 10^{- 19}} \\ = 2.54 - 0.38 = 2.16 e V. \end{array}$

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