KCET · Physics · Atomic Physics
When an electron jumps from the orbit \(n=2\) to \(n=4\), then wavelength of the radiations absorbed will be ( \(R\) is Rydberg's constant)
- A \(\frac{16}{3 R}\)
- B \(\frac{16}{5 R}\)
- C \(\frac{5 R}{16}\)
- D \(\frac{3 R}{16}\)
Answer & Solution
Correct Answer
(A) \(\frac{16}{3 R}\)
Step-by-step Solution
Detailed explanation
Wavelength,
\[
\begin{aligned}
&\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] \\
&\frac{1}{\lambda}=R\left[\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right] \\
&\frac{1}{\lambda}=R\left[\frac{1}{4}-\frac{1}{16}\right] \\
&\frac{1}{\lambda}=R\left[\frac{4-1}{16}\right] \\
&\frac{1}{\lambda}=\frac{3 R}{16} \\
&\lambda=\frac{16}{3 R}
\end{aligned}
\]
\[
\begin{aligned}
&\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] \\
&\frac{1}{\lambda}=R\left[\frac{1}{(2)^{2}}-\frac{1}{(4)^{2}}\right] \\
&\frac{1}{\lambda}=R\left[\frac{1}{4}-\frac{1}{16}\right] \\
&\frac{1}{\lambda}=R\left[\frac{4-1}{16}\right] \\
&\frac{1}{\lambda}=\frac{3 R}{16} \\
&\lambda=\frac{16}{3 R}
\end{aligned}
\]
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