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GUJCET · Physics · Atoms
What is the shortest wavelength present in the Paschen series of spectral lines？
- A 840 nm
- B 320 nm
- C 720 nm
- D 820 nm
Answer & Solution
Correct Answer
(D) 820 nm
Step-by-step Solution
Detailed explanation
(D) 820 nm
for Paschen series, from \(\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{3^{2}}-\frac{1}{n^{2}}\right]\)
\(
n=4,5,6 \ldots
\)
for shortest wavelength taking \(n=\infty\),
\(
\begin{aligned}
\frac{1}{\lambda_{\min }} & =\mathrm{R}\left[\frac{1}{3^{2}}-\frac{1}{\infty^{2}}\right] \\
\lambda_{\min } & =\frac{9}{\mathrm{R}} \\
\therefore \lambda_{\min } & =\frac{9}{1.097 \times 10^{-7}} \\
\therefore \lambda_{\min } & \approx 820.41 \times 10^{-9} \mathrm{~m} \\
\therefore \lambda_{\min } & \approx 820 \mathrm{~nm}
\end{aligned}
\)
for Paschen series, from \(\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{3^{2}}-\frac{1}{n^{2}}\right]\)
\(
n=4,5,6 \ldots
\)
for shortest wavelength taking \(n=\infty\),
\(
\begin{aligned}
\frac{1}{\lambda_{\min }} & =\mathrm{R}\left[\frac{1}{3^{2}}-\frac{1}{\infty^{2}}\right] \\
\lambda_{\min } & =\frac{9}{\mathrm{R}} \\
\therefore \lambda_{\min } & =\frac{9}{1.097 \times 10^{-7}} \\
\therefore \lambda_{\min } & \approx 820.41 \times 10^{-9} \mathrm{~m} \\
\therefore \lambda_{\min } & \approx 820 \mathrm{~nm}
\end{aligned}
\)
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