Question

Light of wavelength strikes a photo-sensitive surface and electrons are ejected with kinetic energy E. If the kinetic energy is to be increased to 2E, the wavelength must be changed to

• A. ${\lambda }^{\prime }=\frac{\lambda }{2}$
• B. ${\lambda }^{\prime }=2\lambda$
• C. $\frac{\lambda }{2}<{\lambda }^{\prime }<\lambda$
• D. ${\lambda }^{\prime }>\lambda$

Answer 1

The correct option is C $\frac{\lambda }{2}<{\lambda }^{\prime }<\lambda$

$E=\frac{hc}{\lambda }-W_{0}and2E=\frac{hc}{{\lambda }^{\prime }}-W_0$
$\Rightarrow \frac{{\lambda }^{\prime }}{\lambda }=\frac{E+W_0}{2E+W_0}\Rightarrow {\lambda }^{\prime }=\lambda \left(\frac{1+W_0/E}{2+W_0/E}\right)$
$Since\frac{(1+W_0/E)}{(2+W_0/E)}>\frac{1}{2}so{\lambda }^{\prime }>\frac{\lambda }{2}$