Question

the energy difference between first excited state and ground state of a given system is 1.96 ev. The transition occurs between first excited state and ground state. Given K= 8.6 X 10^-5 eV/K. The ratio of rates of stimulated emission and spontaneous emission at 300 degree celsius is

Answer 1

Answer:

The energy difference between the first excited state and ground state of a  system is $1.96eV$, the ratio of rates of stimulated emission and spontaneous emission at $300^0C$ is   $\frac{R_{st}}{R_{sp}}=\frac{1}{7\cdot 6}$

Explanation:

When an atom is excited to a higher state from a level$E_{1}$ by absorbing energy which is equal to the energy difference between the two states, sometimes from a higher level it emits photons either by spontaneous emission or by stimulated emission.

we have the rate of stimulated emission is $R_{st}=B_{21}u\left(v\right)N_2$

where $B_{21}$ is the Einstein coefficient, it gives the probability of stimulated emission, $u\left(v\right)$ is the energy density of incident light, and  $N_2$ is the population in level 2 as shown in the figure.

The rate of spontaneous emission is  $R_{sp}=A_{21}N_2$

where$A_{21}$ is the Einstein coefficient, it gives the probability of stimulated emission.

The ratio of stimulated emission to spontaneous emission can be written as

$\frac{Rst}{R_{sp}}=\frac{B_{21}u\left(\nu \right)}{A_{21}}$

According to Planck's radiation formula  

$U\left(\nu \right)=\frac{8\pi h\nu ^3}{c^3}\begin{aligned}\frac{1}{e^{\frac{h\nu }{kT}}-1}\end{aligned}$

$\frac{A_{21}}{B_{21}}=\frac{8\pi h\nu ^3}{c^3}$

$\frac{R_{st}}{Rsp}=\frac{1}{e^{h\nu /kT}-1}$

From the question we have

$E_2-E_1=h\nu E_2-E_1=h\nu =1.96eV$

$\begin{aligned}K=8\cdot 6\times 10^{-5}\frac{ev}{k}\end{aligned}$  

$T=300^{\circ }C=573k$

Substituting these values, we get,

$\frac{1}{e^{\left(\frac{1.96}{8.6\times \overline{10}^5\times 573}\right)_{ }}-1}$

$\frac{1}{2\cdot 17^{39}-1}$ (value of exponential is $e=2.71$)

$\frac{R_{st}}{R_{sp}}=\frac{1}{7\cdot 6}$