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An electron jumps from the $4^{\text {th }}$ orbit to the $2^{\text {nd }}$ orbit of hydrogen atoms. Given the Rydberg's constant $\mathrm{R}_{\mathrm{H}}=10^7 \mathrm{~m}^{-1}$.
frequency in $\mathrm{Hz}$ of the emitted radiation is $\left(\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right.$ )
Options:
frequency in $\mathrm{Hz}$ of the emitted radiation is $\left(\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right.$ )
Solution:
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Verified Answer
The correct answer is:
$\frac{9}{16} \times 10^{15}$
The correct option is (D).
On plugging equation (1) in equation (2),
$\mathrm{f}=\mathrm{cR}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
Given, $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, \mathrm{R}_{\mathrm{H}}=10^7 \mathrm{~m}^{-1}, \mathrm{n}_1=2$ and $\mathrm{n}_2=4$.
$\mathrm{f}=3 \times 10^8 \mathrm{~ms}^{-1} \times 10^7 \mathrm{~m}^{-1}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$
On solving,
$\mathrm{f}=\frac{9}{16} \times 10^{15} \mathrm{~Hz}$
On plugging equation (1) in equation (2),
$\mathrm{f}=\mathrm{cR}_{\mathrm{H}}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$
Given, $\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}, \mathrm{R}_{\mathrm{H}}=10^7 \mathrm{~m}^{-1}, \mathrm{n}_1=2$ and $\mathrm{n}_2=4$.
$\mathrm{f}=3 \times 10^8 \mathrm{~ms}^{-1} \times 10^7 \mathrm{~m}^{-1}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$
On solving,
$\mathrm{f}=\frac{9}{16} \times 10^{15} \mathrm{~Hz}$
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