Question

Monochromatic radiation emitted when electron on hydrogen atom jumps from first excited to the ground state irradiates a photosensitive material. The stopping potential is measured to be $3.57V$. The threshold frequency of the material is

• A. $2.5\times {10}^{15}\text{Hz}$
• B. $1.6\times {10}^{15}\text{Hz}$
• C. $5\times {10}^{15}\text{Hz}$
• D. $4\times {10}^{15}\text{Hz}$

Answer 1

The correct option is B $1.6\times {10}^{15}\text{Hz}$

For hydrogen atom the energy is given by
$E_n=-\frac{13.6}{n^2}\text{eV}$
For ground state, $n=1$
$\therefore E_1=-\frac{13.6}{1^2}=-13.6\text{eV}$

For first excited state, $n=2$
$\therefore E_2=-\frac{13.6}{2^2}=-3.4\text{eV}$

The energy of the emitted photon when an electron jumps from first excited state to ground state is
$h\nu =E_2-E_1=-3.4\text{eV}-(-13.6\text{eV})=10.2\text{eV}$

Maximum kinetic energy,
$K_{max}=e{V}_s=e\times 3.57\text{V}=3.57\text{eV}$

According to Einstein's photoelectric equation
$K_{max}=h\nu -{\varphi }_0$
where ${\varphi }_0$ is the work function and $h\nu$ is the incident energy
${\varphi }_0=h\nu -K_{max}=10.2\text{eV}-3.57\text{eV}=6.63\text{eV}$

Threshold frequency, ${\nu }_0=\frac{{\varphi }_0}{h}=\frac{6.63\times 1.6\times {10}^{-19}\text{J}}{6.63\times {10}^{-34}\text{Js}}=1.6\times {10}^{15}\text{Hz}$

Hence, option $\left(C\right)$ is correct.