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JEE Advance 2011
Paper-1
Question
The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561 \mathrm{~A}^0$. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is
Select the correct option:
A
$1215 \mathrm{~A}^0$
B
$1640 \mathrm{~A}^0$
C
$2430 \mathrm{~A}^0$
D
$4687 \mathrm{~A}^0$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$ \begin{aligned} & \frac{1}{6561}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 \mathrm{R}}{36} \\ & \frac{1}{\lambda}=4 \mathrm{R}\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{3 \mathrm{R} \times 4}{16} \\ & \lambda=1215 \mathrm{~A}^0 \end{aligned} $$
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