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प्रश्न
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
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उत्तर
It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.
When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.
Orbital energy is related to orbit level (n) as:
E = `(-13.6)/(n)^2 ev`
For n = 3, E = `(-13.6)/9` = −1.5 eV
This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.
During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.
We have the relation for wave number for the Lyman series as:
`1/lambda = R_y (1/1^2 - 1/n^2)`
Where
Ry = Rydberg constant = 1.097 × 107 m−1
λ = Wavelength of radiation emitted by the transition of the electron
For n = 3, we can obtain λ as:
`1/lambda = 1.0997 xx 10^7 (1/1^2 - 1/3^2)`
= `1.097 xx 10^7 (1 - 1/9)`
= `1.097 xx 10^7 xx 8/9`
`lambda = 9/(8 xx 1.097 xx 10^7)`
= 102.55 nm
If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:
`1/lambda = 1.097 xx 10^7 (1/1^2 - 1/2^2)`
= `1.097 xx 10^7(1- 1/4)`
= `1.097 xx 10^7 xx 3/4`
`lambda = 4/(1.097 xx 10^7 xx 3)`
= 121.54 nm
If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:
`1/lambda = 1.097 xx 10^7 (1/2^2 - 1/3^2)`
`= 1.097 xx 10^7 (1/4 - 1/9)`
= `1.097 xx 10^7 xx 5/36`
= `36 /(5xx1.097 xx 10^7)`
= 656.33 nm
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence, in the Lyman series, two wavelengths, i.e., 102.55 nm and 121.54 nm, are emitted. And in the Balmer series, one wavelength, i.e., 656.33 nm, is emitted.
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