Question

The kinetic energy of the electron emitted when light of frequency $3.5 \times10^{15}$ Hz falls on a metal surface having threshold frequency $1.5 \times 10^{15}$ Hz is ($h = 6.6 \times 10^{-34}$ Js

• A. $1.32 \times 10^{-18}$J
• B. $3.3 \times 10^{-18}$J
• C. $6.6 \times 10^{-19}$ J
• D. $1.98 \times 10^{-19}$J

Answer 1

Answer: (A)

$KE=h \left(v-v_{o}\right)$
$= 6.6 \times10^{-34} \left(3.5 \times10^{15}-1.5\times10^{15}\right)$
$=1.32 \times10 ^{-18} J$