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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.
संबंधित प्रश्न
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, a thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10−10 m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.
(b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.
If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?
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Let En = `(-1)/(8ε_0^2) (me^4)/(n^2h^2)` be the energy of the nth level of H-atom. If all the H-atoms are in the ground state and radiation of frequency (E2 - E1)/h falls on it ______.
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- some of atoms will move to the first excited state.
- all atoms will be excited to the n = 2 state.
- no atoms will make a transition to the n = 3 state.
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