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Atoms — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsAtoms5 Marks
Question
Consider an excited hydrogen atom in state $n$ moving with a velocity $u(ν << c)$. It emits a photon in the direction of its motion and changes its state to a lower state $m$. Apply momentum and energy conservation principles to calculate the frequency $ν$ of the emitted radiation. Compare this with the frequency $ν_0$ emitted if the atom were at rest.
✓

Answer

Velocity of hydrogen atom in state $‘n’ = u$
Also the velocity of photon $= u$
But $u << C$
Here the photon is emitted as a wave.
So its velocity is same as that of hydrogen atom i.e. $u$.
$\therefore$ Accounding to Doppler’s effect,
Frequency $\text{v}=\text{v}_0\bigg(\frac{1+\frac{\text{u}}{\text{c}}}{1-\frac{\text{u}}{\text{c}}}\bigg)$
as $u <<< C$
$1-\frac{\text{u}}{\text{c}}=\text{q}$
$\therefore\ \text{v}=\text{v}_0\bigg(\frac{1+\frac{\text{u}}{\text{c}}}{1}\bigg)=\text{v}_0\Big(1+\frac{\text{u}}{\text{c}}\Big)$
$\text{v}=\text{v}_0\Big(1+\frac{\text{u}}{\text{c}}\Big)$

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