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Find the Temperature at Which the Average Thermal Kinetic Energy is Equal to the Energy Needed to Take a Hydrogen Atom from Its Ground State to N = 3 State. Hydrogen Can Now

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Question

Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n = 3 state. Hydrogen can now emit red light of wavelength 653.1 nm. Because of Maxwellian distribution of speeds, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.

Sum
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Solution

Given:

Wavelength of red light, λ  = 653.1 nm = 653.1 × 10  m

Kinetic energy of H2 molecules (K) is given by

`K = 3/2 KT` .......(1)

Here, `K = 8.62 xx 10^-5   eV/K`

T = Temperature of H2 molecules

Energy released (E) when atom goes from

ground state to n = 3 is given by

`E = 13.6 (1/n_1^2 - 1/n_2^2)`

For ground state, n1 = 1

Also, n2 = 3  

`therefore E = 13.6 (1/1 - 1/9)`

= `13.6 (8/9)` .............(2)

 kinetic energy of H2 molecules = Energy released when hydrogen atom goes from ground state to n = 3 state

`therefore 3/2 xx 8.62 xx 10^-5xxT = (13.6xx8)/9`

`rArr T = (13.6xx8xx2)/(9xx3xx8.62xx10^-5)` 

= 9.4 × 104  K

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Chapter 43: Bohr’s Model and Physics of Atom - Exercises [Page 385]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 43 Bohr’s Model and Physics of Atom
Exercises | Q 23 | Page 385

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