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Physics Question on Electromagnetic Spectrum

When a hydrogen atom emits a photon during a transition from n=4n = 4 to n=2n = 2, its recoil speed is about

A

4.28 ms−14.28\, ms^{-1}

B

0.814 ms−10.814\, ms^{-1}

C

2.07 ms−12.07\, ms^{-1}

D

0.407 ms−10.407\, ms^{-1}

Answer

0.814 ms−10.814\, ms^{-1}

Explanation

Solution

We know that momentum of a photon, p=hλp=\frac{h}{\lambda}
⇒mv=hλ⇒v=hmλ\Rightarrow m v=\frac{h}{\lambda} \Rightarrow v=\frac{h}{m \lambda} ..(i)
[∵p=mv][\because p=m v]
Given, hydrogen atom emits a photon during a transition from n=4n=4 to n=2n=2, According to Bohr's formula for HH -atom, the wavelength of the emitted photon is given by
∴1λ=R(1n12−1n22)\therefore \frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)
where, R=R= Rydberg's constant =1.097×107m−1=1.097 \times 10^{7} m ^{-1}
⇒1λ=1.097×107(122−142)\Rightarrow \frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)
⇒λ=4.86×10−7m\Rightarrow \lambda=4.86 \times 10^{-7} m
∵\because Mass of a proton =1.6×10−27kg=1.6 \times 10^{-27} kg
Now, from E (i), we get
Hence, recoil speed of photon,
v=6.62×10−341.6×10−27×4.86×10−7v = \frac{6.62 \times 10^{-34}}{1.6 \times 10^{-27} \times 4.86 \times 10^{-7}}
=0.8513 m/s= 0.8513\,m/s
or v≈0.814 ms−1v \approx 0.814\,ms^{-1}