Question

The maximum kinetic energy of photoelectrons emitted from a metal surface increases from $0.4\,eV$ to $1.2\,eV$ 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

According to Einstein's photoelectric equation,
$K_{\max}=h\upsilon-\phi_{0}$
where $K_{\max}$ is the maximum kinetic energy of photoelectrons and $\phi_{0}$ is the work function of the metal surface
Initially,
$h\upsilon=0.4+\phi_{0}$
When the frequency is increased by $40\%$ , then
$1.4h\upsilon=1.2+\phi_{0}$
On solving we get
$\phi_{0}=1.6\,eV$