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प्रश्न
Consider an excited hydrogen atom in state n moving with a velocity υ(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 ν0 emitted if the atom were at rest.
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उत्तर
Let the frequency emitted by the atom at rest be ν0.
Let the velocity of hydrogen atom in state 'n' be u.
But u << c
Here, the velocity of the emitted photon must be u.
According to the Doppler's effect,
The frequency of the emitted radiation, ν is given by
Frequency of the emitted radiation, `v = v_0 ((1 + u/c)/(1 - u/c))`
since u <<< c ,
`v = v_0 ((1+u/c)/1)`
`v = v_0 (1 + u/c)`
Ratio of frequencies of the emitted radiation,
`v/v_0 = (1 + u/e)`
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