Light is incident on the cathode of a photocell and the stopping voltages are measured for light of two different wavelengths. From the data given below, determine the work functions of the metal of the cathode in eV.
Wavelength $(Å)$ Stopping voltage (volt)
4000 1.3
500 0.9

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4.6 eV

2.3 eV

3.2 eV

1.5 eV

answer is B.

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Detailed Solution

As $K_{max} = hv - W_{0}$

or ${eV}_{0} = \frac{hc}{λ} - W$

or $V_{0} = \frac{hc}{eλ} - \frac{W_{0}}{e}$

$\begin{array}{l} {ΔV}_{0} = {(V_{0})}_{2} - {(V_{0})}_{1} = [\frac{hc}{{eλ}_{2}} - \frac{W_{0}}{e}] - [\frac{hc}{{eλ}_{1}} - \frac{W_{0}}{e}] = \frac{hc}{e} [\frac{1}{λ_{2}} - \frac{1}{λ_{1}}] \\ {ΔV}_{0} = \frac{hc}{e} [\frac{λ_{1} - λ_{2}}{λ_{1} λ_{2}}] \\ \frac{hc}{e} = \frac{{ΔV}_{0} \cdot λ_{1} λ_{2}}{λ_{1} - λ_{2}} = \frac{(1 .3 - 0 .9) \times 4000 \times 10^{- 10} \times 4500 \times 10^{- 10}}{500 \times 10^{- 10}} \\ = 1 .44 \times 10^{- 6} Vm \end{array}$

Also, $V_{0} = \frac{hc}{eλ} - \frac{W_{0}}{e}$

$\begin{array}{l} ∴ \frac{W_{0}}{e} = \frac{hc}{eλ} - V_{0} = \frac{1 .44 \times 10^{- 6}}{4000 \times 10^{- 10}} - 1 .3 = 2 .3 V \\ \Rightarrow W_{0} = 2 .3 eV \end{array}$