Question

If the series limit wavelength of the Lyman series for hydrogen atom is \[912~\overset{\circ }{\mathop{\text{A}}}\,\], then the series limit wavelength for the Balmer series for the hydrogen atom is
A. \[912~\overset{\circ }{\mathop{\text{A}}}\,/2\]
B. \[912\overset{\circ }{\mathop{\text{A}}}\,\]
C. \[912\times 2\overset{\circ }{\mathop{\text{A}}}\,\]
D. \[912\times 4\overset{\circ }{\mathop{\text{A}}}\,\]

Answer 1

Hint: Lyman series are the spectral series of transitions from higher orbits to the orbit \[n=1\]. Balmer series are the transitions from higher orbits to the orbit \[n=2\]. Use Rydberg formula to find the wavelength of a particular series of hydrogen atoms.

Complete step by step answer:
Here we are considering the Lyman and Balmer series. These are the specific spectral series of emission spectrum of hydrogen atom.
According to the Rydberg formula,
\[\dfrac{1}{\lambda }=R\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]\]…………..(1)
where \[\lambda \] is the wavelength, \[R\] is Rydberg’s constant. \[{{n}_{1}}\] and \[{{n}_{2}}\] are the initial and final energy levels.
\[\dfrac{1}{\lambda }\] is collectively known as the wave number.
For the limiting wavelength of the Lyman series, we are taking \[{{n}_{2}}\] as \[\infty \].
For Lyman series, \[{{n}_{1}}=1\]
Wavelength of these series are already given in the question, \[\lambda =912\text{ }\overset{\text{o}}{\mathop{\text{A}}}\,\]
We can assign these values to the Rydberg formula.
\[\dfrac{1}{912}=R\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right]\], Rydberg’s constant, \[R=1.097\times {{10}^{7}}{{m}^{-1}}\]
\[\dfrac{1}{R}=912\text{ }\overset{\circ }{\mathop{\text{A}}}\,\]……………(2)
Next, we can find the limiting wavelength of the Balmer series.
For Balmer series, \[{{n}_{1}}=2\] and \[{{n}_{2}}=\infty \]
We can assign these values into Rydberg's equation.
\[\dfrac{1}{\lambda }=R\left[ \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right]\]
\[\dfrac{1}{\lambda }=\dfrac{R}{4}\]
We can alter this equation to compare with equation (2)
\[\dfrac{4}{R}=\lambda \]
We have already found \[\dfrac{1}{R}=912\text{ }\overset{\circ }{\mathop{\text{A}}}\,\]. Thus, we have to multiply it with 4 to get the limiting wavelength of the Balmer series.
\[\dfrac{4}{R}=4\times 912\text{ }\overset{\circ }{\mathop{\text{A}}}\,\]
\[\lambda =3648\text{ }\overset{\text{o}}{\mathop{\text{A}}}\,\]
So, the option D is correct.

Additional information:
Lyman series are the spectral lines emitted by the transitions of electrons from an outer orbit of quantum number \[n>1\] to the first orbit of quantum number \[n=1\].
Balmer series are the transitions from an outer orbit of the quantum number \[n>2\] to the orbit \[n=2\].
Similarly, in the increasing order of quantum numbers, hydrogen have Paschen, Brackett, Pfund, Humphreys series respectively.

Note: Each of the spectral series has its own specific quantum number. If you interchange those you will get wrong answers. For this question you don’t have to multiply with Rydberg’s constant. You can keep like the solution. It will be easy to compare if it is not multiplied with Rydberg’s constant.