Q100 / 160 · JIPMER 2005 · Physics

If work function of a metal is 4.2 eV , the cut off wavelength is

  • A
    $8000 Å$
  • B
    $7000 Å$
  • C
    $1472 Å$
  • Correct answer
    D
    $2950 Å$

Solution

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

Hence, the correct answer is option: 'D'.

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