Question

If the frequency of incident light falling on a photosensitive metal is doubled, the kinetic energy of the emitted photoelectron

• A. unchanged
• B. halved
• C. doubled
• D. more than twice its initial value
• E. reduced to $\frac{1}{4}th$

Answer 1

Answer: (D)

more than twice its initial value

By Einstein's photoelectric equation, The kinetic energy of photoelectrons
$KE=hv-\varphi$
where $v$ = frequency of incidence light
New kinetic energy $K{E}^{\prime }=2hv-\varphi$
So, $K{E}^{\prime }=2hv-2\varphi +\varphi =2KE+\varphi$
So, the new kinetic energy is more than twice its initial value.