Question

At what temperature the rate of spontaneous and stimulated emission are equal . Assume that the wavelength is 500nm.?

Answer 1

Answer:

For an emission of wavelength 5000 Angstrom, both spontaneous and stimulated emissions will be same at temperature 28869.6 K

The Temperature is calculated by the formula

=e^(-hc/λKT)-1

here, n is number of spontaneous emission, N number of stimulated emission, h is Planck's constant,  c is speed of light, K is Boltzmann constant, and T is temperature

Plugging the values in the above equation

1=e^((-6.64*10^(-34)*3*10^8)/(5000*10^(-10)*1.38*(10^-23)*T)

T=28869.6 K

Explanation:

Answer 2

Given : Rate of spontaneous and stimulated emission are equal.

           Wavelength is 500nm.    

To Find : Temperature

Explanation:

The Temperature is calculated by the formula,

The relative rate at which spontaneous and stimulated emission occur in equilibrium at temperature T is

  $\[R = \frac{{{A_{21}}}}{{{B_{21}}\rho \left( \nu \right)}} = 1\]$

The spontaneous emission rate is $\[{{A_{21}}}\]$ which is independent of external radiation $\[{B_{21}}\]$ is called the Einstein coefficient for stimulated emission.

We choose the $\[\rho \left( \nu \right)\]$ as the appropriate for black body radiation field.

$\[R = \frac{{{A_{21}}}}{{{B_{21}}\rho \left( \nu \right)}} = {e^{hv/kt}} - 1 = 1\]$

here,

 h is Planck's constant,

 c is speed of light,

 K is Boltzmann constant,

 T is temperature

now solving it,

$\[\frac{{hv}}{{kT}} = \ln 2\]$

T=41562 K

Hence, Temperature the rate of spontaneous and stimulated emission are equal IS 41562 K.