Question

Light of wavelength $\lambda$, strikes a photoelectric surface and electrons are ejected with an energy E. If E is to be increased to exactly twice its original value, the wavelength changes to ${\lambda }^{\prime }$ where

• A. ${\lambda }^{\prime }$ is less than $\frac{\lambda }{2}$
• B. ${\lambda }^{\prime }$ is greater than $\frac{\lambda }{2}$
• C. ${\lambda }^{\prime }$ is greater than $\frac{\lambda }{2}$but less than$\lambda$
• D. ${\lambda }^{\prime }$ is exactly equal to$\frac{\lambda }{2}$

Answer 1

The correct option is C ${\lambda }^{\prime }$ is greater than $\frac{\lambda }{2}$but less than$\lambda$

$E_{total}=workfunction+KE$

$\frac{hc}{\lambda }=\varphi +E$

now KE become 2E, work function will remain same

$\frac{hc}{{\lambda }^{\prime }}=\varphi +2E$

in order to double E, ${\lambda }^{\prime }$ should be greater than $\frac{\lambda }{2}$ and less than $\lambda$