Question

Determine the ratio of population of to energy level in a medium in thermal equilibrium . if the wave length of light emitted at 330k is 632 nm

Answer 1

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Explanation:

the ratio of population of two energy levels out of which upper one corresponds to a metastabe state is 1.059x10^-30. find the wavelength at 330K

Answer 2

The wavelength of light emitted at 330 K can be found by using the Boltzmann distribution law to determine the energy difference between two energy levels, and then using that energy difference and the wavelength of light to determine the wavelength of the light emitted.

To find the wavelength of light emitted at 330 K when the ratio of population of two energy levels is 1.059 x 10^-30, we can use the Boltzmann distribution law, which relates the ratio of populations of two energy levels to the exponential of the difference in energy between the levels, divided by the temperature in Kelvin.

Assuming that the upper energy level corresponds to a metastable state, the Boltzmann distribution can be written as:

ratio = (N2/N1) = e^((E1 - E2)/kT)

Where N1 and N2 are the populations of the lower and upper energy levels, E1 and E2 are their energies, k is the Boltzmann constant, and T is the temperature in Kelvin.

Rearranging this equation, we can find the energy difference between the two levels:

(E1 - E2) = kT * ln(N2/N1) = kT * ln(ratio)

Next, the energy of each level can be used to find the corresponding wavelength of light emitted using the equation E = hc/λ, where h is Planck's constant, and c is the speed of light.

Finally, we can use the energy difference and the wavelength of light to determine the wavelength of the light emitted at 330 K.

For more such questions on energy: https://brainly.in/question/12756577

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