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When a Photon is Emitted from an Atom, the Atom Recoils. the Kinetic Energy of Recoil and the Energy of the Photon Come from the Difference in Energies Between the States - Physics

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प्रश्न

When a photon is emitted from an atom, the atom recoils. The kinetic energy of recoil and the energy of the photon come from the difference in energies between the states involved in the transition. Suppose, a hydrogen atom changes its state from n = 3 to n = 2. Calculate the fractional change in the wavelength of light emitted, due to the recoil.

बेरीज
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उत्तर

Difference in energy in the transition from n = 3 to n = 2 is 1.89 eV ( = E).

If all this energy is used up in emitting a photon (i.e. recoil energy is zero).

Then,

`E = (hc)/lamda`

`rArr lamda = (hc)/E` ..........(i)

If difference of energy is used up in emitting a photon and recoil of atom, then let ER be the recoil energy of atom.

`E = (hc)/lamda + E_R`

`rArr lamda ' = (hc)/(E - E_R)` ............(ii)

Fractional change in the wavelength is given as,

`(Delta lamda)/lamda = (lamda'-lamda)/lamda`

`rArr (Deltalamda)/lamda = 1/lamda((hc)/(E-E_R ) - (hc)/E)`

`rArr (Delta lamda)/(lamda)=E/(hc)  (hcE_R)/(E(E-E_R))    (therefore lamda = (hc)/E)`

`rArr (Deltalamda)/lamda = ((E_R)/(E- E_R))`

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पाठ 21: Bohr’s Model and Physics of Atom - Exercises [पृष्ठ ३८५]

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एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
पाठ 21 Bohr’s Model and Physics of Atom
Exercises | Q 37 | पृष्ठ ३८५

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.


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