MHT CET · Physics · Atomic Physics
According to Bohr's theory of hydrogen atom, the ratio of the maximum and minimum wavelength of Lyman series will be
- A \(3: 4\)
- B \(4: 3\)
- C \(2: 5\)
- D \(5: 2\)
Answer & Solution
Correct Answer
(B) \(4: 3\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\)
For Lyman series, \(\mathrm{n}_1=1\)
\(\begin{aligned}
\quad \frac{1}{\lambda_{\max }} & =R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4} R \\
\frac{1}{\lambda_{\min }} & =R\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)=R \\
\therefore \quad & \frac{\lambda_{\max }}{\lambda_{\min }}
\end{aligned}=\frac{4}{3}\)
For Lyman series, \(\mathrm{n}_1=1\)
\(\begin{aligned}
\quad \frac{1}{\lambda_{\max }} & =R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3}{4} R \\
\frac{1}{\lambda_{\min }} & =R\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)=R \\
\therefore \quad & \frac{\lambda_{\max }}{\lambda_{\min }}
\end{aligned}=\frac{4}{3}\)
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