Q148 / 160 · AP EAMCET 2003 · Physics

When radiation of the wavelength $\lambda$ is incident on a metallic surface, the stopping potential is $4.8 \mathrm{~V}$. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes $1.6 \mathrm{~V}$. Then, the threshold wavelength for the surface is :

  • A
    $2 \lambda$
  • Correct answer
    B
    $4 \lambda$
  • C
    $6 \lambda$
  • D
    $8 \lambda$

Solution

Stopping potential. $V_0=\frac{h c}{e}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)$ where, $V_0=$ stopping potential Ist case, $4.8=\frac{h c}{e}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)$ IInd case, $1.6=\frac{h c}{e}\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right)$ Dividing Eqs. (i) by (ii), $3=\frac{\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}\right)}{\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}\right)}$ $\frac{3}{2 \lambda}-\frac{3}{\lambda_0}=\frac{1}{\lambda}-\frac{1}{\lambda_0}$ $\frac{1}{\lambda_0}-\frac{3}{\lambda_0}=\frac{1}{\lambda}-\frac{3}{2 \lambda}$ $\frac{-2}{\lambda_0}=\frac{2-3}{2 \lambda}$ $\frac{2}{\lambda_0}=\frac{1}{2 \lambda}$ $\lambda_0=4 \lambda$

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

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