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Let $\lambda_P$ and $\lambda_L$ be the longest wavelengths observed in the Paschen and Lyman series respectively. Choose the correct option
Options:
Solution:
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Verified Answer
The correct answer is:
$15 < \frac{\lambda_\rho}{\lambda_L} < 16$
As we know, wavelength for Paschen-series.
$\frac{1}{\lambda_P}=R\left[\frac{1}{3^2}-\frac{1}{n_p^2}\right], n_p=4,5,6$
and wavelength for Lyman-series,
$\frac{1}{\lambda_L}=R\left[1-\frac{1}{n_L^2}\right], n_L=2,3,4 \ldots$
Hence, $\lambda_p$ for $n_p=4$
$\frac{1}{\lambda_p}=R\left[\frac{1}{9}-\frac{1}{16}\right],=\frac{7 R}{9 \times 16}$
and $\lambda_L$ for $n_L=2$
$\frac{1}{\lambda_L}=R\left[1-\frac{1}{4}\right]=\frac{3 R}{4}$
From Eq. (i) and (ii), we get
$\frac{\lambda_p}{\lambda_L}=\frac{3 R / 4}{7 R / 9 \times 16}=15.42$
So, $15 < \frac{\lambda_P}{\lambda_L} < 16$
Hence, the correct option is (c).
$\frac{1}{\lambda_P}=R\left[\frac{1}{3^2}-\frac{1}{n_p^2}\right], n_p=4,5,6$
and wavelength for Lyman-series,
$\frac{1}{\lambda_L}=R\left[1-\frac{1}{n_L^2}\right], n_L=2,3,4 \ldots$
Hence, $\lambda_p$ for $n_p=4$
$\frac{1}{\lambda_p}=R\left[\frac{1}{9}-\frac{1}{16}\right],=\frac{7 R}{9 \times 16}$
and $\lambda_L$ for $n_L=2$
$\frac{1}{\lambda_L}=R\left[1-\frac{1}{4}\right]=\frac{3 R}{4}$
From Eq. (i) and (ii), we get
$\frac{\lambda_p}{\lambda_L}=\frac{3 R / 4}{7 R / 9 \times 16}=15.42$
So, $15 < \frac{\lambda_P}{\lambda_L} < 16$
Hence, the correct option is (c).
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