Question

A photon of energy $8.6\text{eV}$ is incident on a metal surface of threshold frequency $1.6\times {10}^{15}\text{Hz}$. The maximum kinetic energy of the photoelectrons emitted (in $\text{eV}$):

• A. $1.2$
• B. $1.6$
• C. $2$
• D. $6$

Answer 1

The correct option is C $2$

As we know,
${\text{KE}}_{\text{max}}=\text{Incident energy - Work function}$
$\Rightarrow {\text{KE}}_{\text{max}}=8.6\text{eV}-h\nu$
$\Rightarrow {\text{KE}}_{\text{max}}=8.6-\frac{6.63\times {10}^{-34}\times 1.6\times {10}^{15}}{1.6\times {10}^{-19}}\text{eV}$
$\Rightarrow {\text{KE}}_{\text{max}}=8.6-6.63$
$\Rightarrow {\text{KE}}_{\text{max}}=8.6-6.63$
$\Rightarrow {\text{KE}}_{\text{max}}\approx 2\text{eV}$