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If work function of a metal is 4.2 eV , the cut off wavelength is
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$2950 Å$
Suppose the cut off wavelength is represented by $\lambda_0$
So, $\frac{h c}{\lambda_0}=$ work function $=W_0$
$\begin{aligned} \frac{h c}{\lambda_0} & =4.2 \times 1.6 \times 10^{-19} \\ \lambda_0 & =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{4.2 \times 1.6 \times 10^{-19}} \\ & =2946 \times 10^{-10} \mathrm{~m} \\ & \approx 2950 Å\end{aligned}$
So, $\frac{h c}{\lambda_0}=$ work function $=W_0$
$\begin{aligned} \frac{h c}{\lambda_0} & =4.2 \times 1.6 \times 10^{-19} \\ \lambda_0 & =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{4.2 \times 1.6 \times 10^{-19}} \\ & =2946 \times 10^{-10} \mathrm{~m} \\ & \approx 2950 Å\end{aligned}$
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