Q. A light whose frequency is equal to $6 \times 10^{14}Hz\ $ is incident on a metal whose work function is 2eV. $\lbrack h = 6.63 \times 10^{- 34}\ Js,\ 1eV = 1.6 \times 10^{- 19}\ J\rbrack.$ 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
Explanation
(c) $KE_{\max} = hv - \phi$
Where $hv = energy\ of\ incident\ photon,$
$\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
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.