Question

When a $1cm$ thick surface is illuminated with light of wavelength $\lambda $ , the stopping potential is $V_{0}$ . When the same surface is illuminated by the light of wavelength $3\lambda $ , the stopping potential is $\frac{V_{0}}{6}$ . The threshold wavelength for the metallic surface is

• A. $4\lambda $
• B. $5\lambda $
• C. $3\lambda $
• D. $2\lambda $

Answer 1

Answer: (B)

Let $\lambda _{0}$ be threshold wavelength,
$eV_{0}=hc\left(\frac{1}{\lambda } - \frac{1}{\left(\lambda \right)_{0}}\right)$
Also,
$e\frac{V_{0}}{6}=hc\left(\frac{1}{3 \lambda } - \frac{1}{\left(\lambda \right)_{0}}\right)$
$\therefore \, \, \, \frac{1}{6}\left(\frac{1}{\lambda } - \frac{1}{\left(\lambda \right)_{0}}\right)=\left(\frac{1}{3 \lambda } - \frac{1}{\left(\lambda \right)_{0}}\right)$
$\frac{1}{\left(\lambda \right)_{0}}\left(1 - \frac{1}{6}\right)=\frac{1}{\lambda }\left(\frac{1}{3} - \frac{1}{6}\right)$
$\frac{5}{6 \lambda _{0}}=\frac{1}{6 \lambda }$
$\lambda _{0}=5\lambda $