Question

Monochromatic radiation emitted when electron on hydrogen atom jumps from first exited to the grounds state irradiates a photosnesitive material. The stopping potential is measured to be 3.57V. The threshold frequency of the material is :

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

Answer 1

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

For hydrogen atom,

$E_n=-\frac{13.6}{n^2}eV$

For ground state, $n=1$

$E_1=-\frac{13.6}{1^2}=-13.6eV$

For first excited state, $n=2$

$E_2=-\frac{13.6}{2^2}=-3.4eV$

The energy of the emitted photon when an electron jumps from first excited state to ground state is,

$h\nu =E_2-E_1$

$=-3.4eV-(-13.6eV)=10.2eV$

Maximum kinetic energy,

$K_{max}=e{V}_s=e\times 3.57V=3.57eV$

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

${\varphi }_0=h\nu -K_{max}$

$=10.2eV-3.57eV=6.63eV$

Threshold frequency,

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