Question

In a photoelectric setup, the radiations from the Balmer series of hydrogen atom are incident on a metal surface of work function $2eV$. The wavelength of incident radiations lies between $450nm$ to $700nm$. Find the maximum kinetic energy of photoelectron emitted. (Given $hc/e=1241eV-nm)$.

• A. 0.45 eV
• B. 0.55 eV
• C. 0.65 eV
• D. 0.75 eV

Answer 1

The correct option is B 0.55 eV

$\text{Given}:$ ${\varphi }_0=2eV$,

$\frac{hc}{e}=1242eV-nm$

$\text{Solution}:$

The wavelength of incident radiation lies between 450nm to 700nm of Balmer series. It arises when transition takes place from $n_2=4$ to $n_1=2$ in case of hydrogen atom.

For the transition, $n_1=2$ and $n_2=n$$ energy emitted is

$\Delta E=13.6(\frac{1}{2^2}-\frac{1}{n^2})=\frac{hc}{e\lambda }=\frac{1242}{\lambda }$ or

$\lambda =\frac{1242\times 4n^2}{13.6(n^2-4)}nm$

When $n=4$, then

$\lambda =\frac{1242\times 4\times 4n^2}{13.6(4^2-4)}=487.05nm$

Maximum K.E. of emitted photoelectron is,

$E_{max}=\frac{hc}{e\lambda }-{\varphi }_0\left(ineV\right)=\frac{1242}{487.05-2}=2.55-2=0.55eV$

$\text{Hence B is the correct option}$