Question

The wavelength of the first spectral line in the Balmer series of the hydrogen atom is $6561\overset{\circ }{\mathop{\text{A}}}\,$ . The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is:
(A) $1215\overset{\circ }{\mathop{\text{A}}}\,$
(B) $1640\overset{\circ }{\mathop{\text{A}}}\,$
(C) $2460\overset{\circ }{\mathop{\text{A}}}\,$
(D) $4687\overset{\circ }{\mathop{\text{A}}}\,$

Answer 1

Hint: The wavelength of the Balmer series can be calculated by the formula $\dfrac{1}{\lambda }=R\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right){{Z}^{2}}$where$\lambda $ is the wavelength of the line, $R$ is the Rydberg constant, ${{n}_{1}}$ is 2 for Blamer series, ${{n}_{2}}$ is 3, 4, 5……..depending on the line, and $Z$ is the atomic number of the element.

Complete step by step solution:
The Balmer series is the second series of the hydrogen emission spectrum. So, the wavelength of the lines can be calculated by the formula:
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right){{Z}^{2}}$
Where $\lambda $is the wavelength of the line, R is the Rydberg constant, ${{n}_{1}}$ is 2 for Blamer series, ${{n}_{2}}$ is 3, 4, 5……..depending on the line, and $Z$ is the atomic number of the element.
Now given in the question, for the first line of the Balmer series, the wavelength of a hydrogen atom is $6561\overset{\circ }{\mathop{\text{A}}}\,$. So,
$\lambda $ = $6561\overset{\circ }{\mathop{\text{A}}}\,$
${{n}_{1}}$ is 2 for Blamer series
${{n}_{2}}$ = 3 (for first line)
$Z$ = 1 (atomic number of a hydrogen atom)
So putting all these in the equation:
$\dfrac{1}{6561}=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right){{1}^{2}}$
$\dfrac{1}{6561}=R\left( \dfrac{5}{36} \right)$
Now given in the question, for the second line of the Balmer series, the wavelength is of helium atom will be,
${{n}_{1}}$ is 2 for Blamer series
${{n}_{2}}$ = 4 (for second line)
$Z$ = 2 (atomic number of a helium atom)
So putting all these in the equation:
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{4}^{2}}} \right){{2}^{2}}$
$\dfrac{1}{\lambda }=R\left( \dfrac{3}{4} \right)$
So by dividing both the equation of the lines, we get:
$\dfrac{\dfrac{1}{6561}=R\left( \dfrac{5}{36} \right)}{\dfrac{1}{\lambda }=R\left( \dfrac{3}{4} \right)}$
So, on solving we get,
$\dfrac{\lambda }{6561}=\dfrac{5}{36}\ \text{x }\dfrac{4}{3}$
$\lambda =\dfrac{5}{36}\ \text{x }\dfrac{4}{3}\text{ x 6561}$
$\lambda =1215\overset{\circ }{\mathop{\text{A}}}\,$
So, the wavelength for the second spectral line will be $1215\overset{\circ }{\mathop{\text{A}}}\,$.

Therefore, the correct answer is an option (A) $1215\overset{\circ }{\mathop{\text{A}}}\,$.

Note: In the question solved above $\overset{\circ }{\mathop{\text{A}}}\,$ is the angstrom unit which is equal to ${{10}^{-10}}$ the meter. The third line in the Balmer series ${{n}_{2}}$ will be 5 and so on. ${{n}_{1}}$ remains constant for each series.