Question

Ultraviolet radiation of $6.2\, eV$ falls on an aluminium foil surface. Work function is $4.2\, eV$. The kinetic energy of the fastest electron emitted approximately

• A. $3.2 \times 10^{-21} J$
• B. $3.2 \times 10^{-19} J$
• C. $3.2 \times 10^{-15} J$
• D. $3.2 \times 10^{-17} J$

Answer 1

Answer: (B)

From Einstein's relation, the kinetic energy $(KE)$ of emitted electron is given by
$KE =h v -W_{0}$
where, $W_{0}$ is work function of metal, $v$ the frequency, $h$ the Planck's constant.
Given, $h v=6.2\, eV,\, W_{0}=4.2\, eV$
$\therefore KE =6.2-4.2=2\, eV$
Also, $1\, eV =1.6 \times 10^{-19} J$
$\therefore KE =2 \times 1.6 \times 10^{-19} J$
$=3.2 \times 10^{-19} J$