Question

When the energy of incident radiation is increased by $20 \%$, the kinetic energy of the photo electrons emitted from a metal surface increased from $0.5 \,eV$ to $0.8\, eV$. The work function of the metal is

• A. $0.65\, eV$
• B. $1.0 \,eV$
• C. $1.3 \,eV$
• D. $1.5 \,eV$

Answer 1

Answer: (B)

According to Einstein's photoelectric equation,
$E_{K}=E-\varphi_{0}$
where, $E_{K}=$ kinetic energy,
$E=$ energy of incident photon
and $\varphi_{0}=$ work function of metal surface.
$\Rightarrow E=E_{K}+\varphi_{0} \ldots$ (i)
When energy of incident photon is increased by $20 \%$, then
$E'=E+20 \%$ of $E=E+\frac{20}{100} E=\frac{6 E}{5}$
$\Rightarrow E'=1.2 E$
$\therefore E_{K}'=E'-\varphi_{0}$
$ \Rightarrow E'=E_{K}'+\varphi_{0}$
$1.2 E=E_{K}'+\varphi_{0} \ldots$ (ii)
From Eqs. (i) and (ii), we get
$\frac{E}{1.2 E}=\frac{E_{K}+\varphi_{0}}{E_{K}{ }'+\varphi_{0}}=\frac{0.5+\varphi_{0}}{0.8+\varphi_{0}}$
$\left[\because E_{K}=0.5 eV\right.$ and $\left.E_{K}{ }'=0.8 eV \right]$
$\Rightarrow \frac{5}{6}=\frac{0.5+\varphi_{0}}{0.8+\varphi_{0}}$
$4+5 \varphi_{0}=3+6 \varphi_{0}$
$ \Rightarrow \varphi_{0}=1 eV$