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If $\mathrm{R}$ is the Rydberg Constant in $\mathrm{cm}^{-1}$, then hydrogen atom does not emit any radiation of wave-length in the range of
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Verified Answer
The correct answer is:
$\frac{7}{5 \mathrm{R}}$ to $\frac{19}{5 \mathrm{R}} \mathrm{cm}$
Hint:
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{\mathrm{f}^{2}}}-\frac{1}{\mathrm{n}_{\mathrm{i}^{2}}}\right]$
For range of wavelengths:$\mathrm{n}_{\mathrm{i}}=1,2,3, \ldots . .$ for Lyman, Balmer, Paschen, $\ldots \ldots$
$\mathrm{n}_{\mathrm{f}}=\mathrm{n}_{\mathrm{i}}+1$ and $\mathrm{n}_{\mathrm{f}}=\infty$ for upper and lower range
Thus, Lyman : $\left[\frac{1}{\mathrm{R}}\right.$ to $\left.\frac{4}{3 \mathrm{R}}\right]$, Balmer : $\left[\frac{4}{\mathrm{R}}\right.$ to $\left.\frac{36}{5 \mathrm{R}}\right]$, Paschen : $\left[\frac{9}{\mathrm{R}} \mathrm{to} \frac{144}{36 \mathrm{R}}\right]$, Bracket : $\left[\frac{16}{\mathrm{R}}\right.$ to $\left.\frac{400}{9 \mathrm{R}}\right]$, Pfund : $\left[\frac{25}{\mathrm{R}}\right.$ to $\left.\frac{900}{11 \mathrm{R}}\right]$
$\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{\mathrm{f}^{2}}}-\frac{1}{\mathrm{n}_{\mathrm{i}^{2}}}\right]$
For range of wavelengths:$\mathrm{n}_{\mathrm{i}}=1,2,3, \ldots . .$ for Lyman, Balmer, Paschen, $\ldots \ldots$
$\mathrm{n}_{\mathrm{f}}=\mathrm{n}_{\mathrm{i}}+1$ and $\mathrm{n}_{\mathrm{f}}=\infty$ for upper and lower range
Thus, Lyman : $\left[\frac{1}{\mathrm{R}}\right.$ to $\left.\frac{4}{3 \mathrm{R}}\right]$, Balmer : $\left[\frac{4}{\mathrm{R}}\right.$ to $\left.\frac{36}{5 \mathrm{R}}\right]$, Paschen : $\left[\frac{9}{\mathrm{R}} \mathrm{to} \frac{144}{36 \mathrm{R}}\right]$, Bracket : $\left[\frac{16}{\mathrm{R}}\right.$ to $\left.\frac{400}{9 \mathrm{R}}\right]$, Pfund : $\left[\frac{25}{\mathrm{R}}\right.$ to $\left.\frac{900}{11 \mathrm{R}}\right]$
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