Question

A light whose frequency is equal to $6 \times 10^{14} \,Hz$ is incident on a metal whose work function is $2 eV .\left[h=6.63 \times 10^{-34} \,Js , 1\, eV =1.6 \times 10^{-19} \,J \right]$ The maximum energy of the electrons emitted will be

• A. 2.49 eV
• B. 4.49 eV
• C. 0.49 eV
• D. 5.49 eV

Answer 1

Answer: (C)

$KE _{\max }=h v-\phi$
where $h v=$ energy of incident photon,
$\phi=$ work function
$KE _{\max }=6.6 \times 10^{-34} \times 6 \times 10^{14}-2 \times 1.6 \times 10^{-19}$
$=3.96 \times 10^{-19}-3.2 \times 10^{-19}$
$=\frac{0.76 \times 10^{-19}}{1.6 \times 10^{-19}}\, eV$
$=0.475\, eV$