Q104 / 120 · JEE Advanced 2006 · Physics
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 $n$.
Correct answer: 24
Solution
$n^{\text {th }}$ line of Lyman series means transition from $(n+1)$ th state to first state.
$$
\begin{aligned}
& \frac{1}{\lambda}=R Z^2\left[1-\frac{1}{(n+1)^2}\right] \\
& \text { de-Broglie wavelength, } \quad \lambda=\frac{h}{m v}=\frac{h r}{m v r}=\frac{(2 \pi)(h r)}{(n+1) h}=\frac{2 \pi r}{(n+1)} \\
& \frac{1}{\lambda}=\frac{(n+1)}{2 \pi r} \\
&
\end{aligned}
$$
or
Equating (i) and (ii), we get
$$
\left(\frac{n+1}{2 \pi r}\right)=R Z^2\left[\frac{n(n+2)}{(n+1)^2}\right]
$$
Now, as
$$
\begin{aligned}
& r \propto \frac{n^2}{Z} \\
& r=\frac{(n+1)^2}{11} r_0
\end{aligned}
$$
$$
\therefore \quad r=\frac{(n+1)^2}{11} r_0
$$
Substituting in equations (iii), we get
$$
\frac{11}{2 \pi r_0}=\frac{R(11)^2(n)(n+2)}{(n+1)}
$$
or
$$
(n+1)=\left(1.09 \times 10^7\right)(11)(2 \pi) \times\left(0.529 \times 10^{-10}\right)\left(n^2+2 n\right)
$$
Solving this equation we get, $n=24$
Hence, the correct answer is 24.