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प्रश्न
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.
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उत्तर
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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