JEE Advanced · Physics · 27. Atomic 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 \(n\).
- A 20
- B 21
- C 22
- D 24
Answer & Solution
Correct Answer
(D) 24
Step-by-step Solution
Detailed explanation
\(n^{\text {th }}\) line of Lyman series means transition from \((n+1)\) th state to first state.
\(\frac{1}{\lambda}=R Z^2\left[1-\frac{1}{(n+1)^2}\right] \)
\( \text { de-Broglie wavelength, } \quad \lambda=\frac{h}{m v}=\frac{h r}{m v r}=\) \(\frac{(2 \pi)(h r)}{(n+1) h}=\frac{2 \pi r}{(n+1)} \)
\( \frac{1}{\lambda}=\frac{(n+1)}{2 \pi r}\)
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, \(n=24\)
\(\frac{1}{\lambda}=R Z^2\left[1-\frac{1}{(n+1)^2}\right] \)
\( \text { de-Broglie wavelength, } \quad \lambda=\frac{h}{m v}=\frac{h r}{m v r}=\) \(\frac{(2 \pi)(h r)}{(n+1) h}=\frac{2 \pi r}{(n+1)} \)
\( \frac{1}{\lambda}=\frac{(n+1)}{2 \pi r}\)
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, \(n=24\)
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