Question

If the work function for a certain metal is $3.2 \times 10^{-19} J$ and it is illuminated with light of frequency $v=8 \times 10^{14} Hz$, the maximum kinetic energy of the photoelectron would be

• A. $2.1 \times 10^{-19} J$
• B. $3.2 \times 10^{-19} J$
• C. $5.3 \times 10^{-19} J$
• D. $8.5 \times 10^{-19} J$

Answer 1

Answer: (A)

Maximum $KE =h v-\phi_{0} $
$=6.63 \times 10^{-34} \times 8 \times 10^{14}-3.2 \times 10^{-19} $
$=2.1 \times 10^{-19} J $