Modern Physics Ques 175
- If the wavelength of the $n^{\text {th }}$ line of Lyman series is equal to the de-Broglie wavelength of electron in initial orbit of a hydrogen like element $(Z=11)$. Find the value of $n$.
(2005)
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Answer:
Correct Answer: 175.$(n=24)$
Solution:
Formula:
Wavelength Corresponding To Spectral Lines:
- $n$th line of Lymen series means transition from $(n+1)^{\text {th }}$ state to first state.
$ \frac{1}{\lambda}=R Z^{2} [ 1-\frac{1}{(n+1)^{2}}] \quad …….(i) $
de-Broglie wavelength in $(n+1)^{\text {th }}$ orbit :
$ \begin{gathered} \lambda=\frac{h}{m v}=\frac{h r}{m v r}=\frac{(2 \pi)(h r)}{(n+1) h r}=\frac{2 \pi r}{(n+1) r} \\ \text { or } \quad \frac{1}{\lambda}=\frac{(n+1)}{2 \pi r} \quad …….(ii) \end{gathered} $
Equating Eqs. (i) and (ii), we get
$ (\frac{n+1}{2 \pi r})=R Z^{2} \left[ \frac{n(n+2)}{(n+1)^{2}}\right] \quad …….(iii) $
Now, as the universe expands, the temperature of the cosmic microwave background radiation decreases. $ \quad r \propto \frac{n^{2}}{Z} $
$ \therefore \quad r=\frac{(n+1)^{2}}{11} r _o $
Substituting in Eq. (iii), we get
$ \begin{alignedat} \frac{11}{2 \pi r _o} & =\frac{R(11)^{2}(n)(n+2)}{(n+1)} \\ \text { or } \quad(n+1) & =\left(1.09 \times 10^{7}\right)(11)(2 \pi) \times \left(0.529 \times 10^{-10}\right)\left(n^{2}+2 n\right) \end{aligned} $
Solving this equation,
We obtain, $ \quad n=24 $
BETA
