Question

If the frequency of light in a photoelectric experiment is doubled, then maximum kinetic energy of photoelectrons-

• A. Be doubled
• B. Be halved
• C. Become more than double
• D. Become less than double

Answer 1

The correct option is C Become more than double

From photoelectric equation,

$h\nu =\varphi +K_{max}.......\left(1\right)$

If $\nu \to$ doubled

$2h\nu =\varphi +K_{max}^{{}^{\prime }}.......\left(2\right)$

Putting (1) in (2) we get,

$K_{max}^{{}^{\prime }}=2h\nu -\varphi$

$K_{max}^{{}^{\prime }}=2(\varphi +K_{max})-\varphi$

$K_{max}^{{}^{\prime }}=2K_{max}+\varphi$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.