Question

The work function of a metal is $4.2 \,eV$. If radiations of $2000 \mathring{A}$ fall on the metal, then the kinetic energy of the fastest photoelectron is:

• A. $1.6 \times 10^{-19} J$
• B. $16 \times 10^{10} J$
• C. $3.2 \times 10^{-19} J$
• D. $6.4 \times 10^{-10} J$

Answer 1

Answer: (C)

$KE =($ Energy of radiation $-$ Work function $)$
$=\left(h \times \frac{c}{\lambda}-4.2\right)$
$=\left(\frac{6.6 \times 10^{-34} Js \times 3 \times 10^8 m }{2000 \times 10^{-10} m }\right)$
$-\left(4.2 \times 1.602 \times 10^{-19} J \right)$
$=\left(9.9 \times 10^{-19} J \right)-\left(6.7 \times 10^{-19} J \right)$
$=3.2 \times 10^{-19} J$