Question 359: Atoms & Nuclei

Question

A hydrogen atom changes its state from $$n=3$$ to $$n=2$$. Due to recoil, the percentage change in the wave length of emitted light is approximately $$1 	imes 10^{-n}$$. The value of $$n$$ is _______. [Given Rhc $$=13.6 \mathrm{~eV}, \mathrm{hc}=1242 \mathrm{~eV} \mathrm{~nm}, \mathrm{h}=6.6 	imes 10^{-34} \mathrm{~J} \mathrm{~s}$$ mass of the hydrogenatom $$=1.6 	imes 10^{-27} \mathrm{~kg}$$]

Accepted Answer

$$\begin{aligned}
\Delta E & =13.6 \mathrm{eV}\left(\frac{1}{4}-\frac{1}{9}ight) \
& =\frac{68}{36} \mathrm{eV}=1.89 \mathrm{eV}
\end{aligned}$$
Due to recoil of hydrogen atom, the energy of emitted photon will decrease by very small amount.
So for approximate calculations,
$$\begin{aligned}
& \% 	ext { charge }= \frac{\Delta E_{	ext {atom }}}{\Delta E} 	imes 100 \
&=\frac{\frac{\left(\frac{\Delta E}{C}ight)^2}{2 m}}{\Delta E} 	imes 100 \
&=\frac{\Delta E}{C^2 	imes 2 m} 	imes 100 \
&=\frac{1.89 	imes 1.6 	imes 10^{-19} 	imes 100}{\left(3 	imes 10^8ight)^2 	imes 2 	imes 1.6 	imes 10^{-27}} \
&=1.05 	imes 10^{-7} \% \
& 	herefore n=7
\end{aligned}$$

Atoms & Nuclei

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