Question

The maximum kinetic energy of photoelectrons emitted from a metal surface increases from $0.4eV$ to $1.2eV$ when the frequency of the incident radiation is increased by $40\%$ . The work function of the metal surface is $\frac{n}{10}eV$ , what is the value of $n$ ?

Answer 1

Answer: 16

According to Einstein's photoelectric equation,
$K_{\max }=h\upsilon -{\varphi }_0$
where $K_{\max }$ is the maximum kinetic energy of photoelectrons and ${\varphi }_0$ is the work function of the metal surface
Initially,
$h\upsilon =0.4+{\varphi }_0$
When the frequency is increased by $40\%$ , then
$1.4h\upsilon =1.2+{\varphi }_0$
On solving we get
${\varphi }_0=1.6eV$