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प्रश्न
A hydrogen atom in a state having a binding energy of 0.85 eV makes transition to a state with excitation energy 10.2 e.V (a) Identify the quantum numbers n of the upper and the lower energy states involved in the transition. (b) Find the wavelength of the emitted radiation.
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उत्तर
(a) The binding energy of hydrogen is given by
`E = 13.6/(n^2)eV`
For binding energy of 0.85 eV,
`n_2^2 = 13.6/0.85`
`n_2 = 4`
For binding energy of 10.2 eV,
`n_2^2 = 13.6/10.2 = 16`
n2 = 1.15
⇒ n1 = 2
The quantum number of the upper and the lower energy state are 4 and 2, respectively.
(b) Wavelength of the emitted radiation (λ) is given by
`1/lamda = R (1/n_1^2 - 1/n_2^2)`
Here,
R = Rydberg constant
n1 and n2 are quantum numbers.
`therefore 1/lamda = 1.097 xx 10^7 (1/4 - 1/16)`
`rArr lamda = (16)/(1.097xx3xx10^7)`
= 4 .8617 × 10-7
= 487 nm
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