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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.
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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