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 $0.5eV$ to $0.8eV$. The work function of the metal is

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

Answer 1

Answer: (B)

According to Einstein's photoelectric equation,
$E_K=E-{\phi }_0$
where, $E_K=$ kinetic energy,
$E=$ energy of incident photon
and ${\phi }_0=$ work function of metal surface.
$\Rightarrow E=E_K+{\phi }_0\dots$ (i)
When energy of incident photon is increased by $20\%$, then
$E^{\prime }=E+20\%$ of $E=E+\frac{20}{100}E=\frac{6E}{5}$
$\Rightarrow E^{\prime }=1.2E$
$\therefore E_K^{\prime }=E^{\prime }-{\phi }_0$
$\Rightarrow E^{\prime }=E_K^{\prime }+{\phi }_0$
$1.2E=E_K^{\prime }+{\phi }_0\dots$ (ii)
From Eqs. (i) and (ii), we get
$\frac{E}{1.2E}=\frac{E_K+{\phi }_0}{E_K{}^{\prime }+{\phi }_0}=\frac{0.5+{\phi }_0}{0.8+{\phi }_0}$
$[\because E_K=0.5eV$ and $E_K{}^{\prime }=0.8eV]$
$\Rightarrow \frac{5}{6}=\frac{0.5+{\phi }_0}{0.8+{\phi }_0}$
$4+5{\phi }_0=3+6{\phi }_0$
$\Rightarrow {\phi }_0=1eV$