Question

Light of wavelength ${}^{\prime }{\lambda }^{\prime }$ strikes a photosensitive 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 ${}^{\prime }{\lambda }^{\prime }$ where

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

Answer 1

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

From first given condition,
$\frac{hC}{\lambda }-W=E\Rightarrow W=\frac{hC}{\lambda }-E$
Now, For Kinetic Energy to increase to 2E,
$\frac{hC}{{\lambda }_1}-\frac{hC}{\lambda }+E=2E$
$\Rightarrow \frac{hC}{{\lambda }_1}=E+\frac{hC}{\lambda }\Rightarrow \frac{\lambda E+hC}{\lambda hC}=\frac{1}{{\lambda }_1}$

$\Rightarrow {\lambda }_1=\lambda \left(\frac{hC}{hC+E}\right)\Rightarrow {\lambda }_1=\lambda \left(\frac{1}{1+\frac{E}{hC}}\right)$ ------------------------- (I)
So, Clearly from (I), we see denominator is greater than 1, so fraction is less than 1.
So, ${\lambda }_1<\lambda$ and ${\lambda }_1>\frac{\lambda }{2}$ as $\frac{E}{hC}$ could never reach 1 as E is after subtraction of $W$.

So, the answer is option (C).