Question

If the series limit wavelength of Lyman series for the hydrogen atom is 912$\overset{0}{\mathop{A}}\,$, then the series limit wavelength for Balmer series of hydrogen atoms is
$\text{A}\text{. }912\overset{0}{\mathop{A}}\,$
$\text{B}\text{. }912\times 2\overset{0}{\mathop{A}}\,$
$\text{C}\text{. }912\times 4\overset{0}{\mathop{A}}\,$
$\text{D}\text{. }\dfrac{912}{2}\overset{0}{\mathop{A}}\,$

Answer 1

Hint: Use the formula of the wavelength of the emission i.e. $\dfrac{1}{\lambda }={{R}_{H}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$. For the series limit wavelength of the Lyman series ${{n}_{1}}=1$ and ${{n}_{2}}=\infty$ and for the Balmer series, ${{n}_{1}}=2$ and ${{n}_{2}}=\infty$

Formula used:
$\dfrac{1}{\lambda }={{R}_{H}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$

Complete step by step answer:
An atom has many imaginary orbits around the nucleus of the atom in which the electron revolves. These orbits are also called shells (K, L, M and so on). The energy of the revolving electrons increases as we go from the ${{1}^{st}}$ shell to the ${{n}^{th}}$ shell. Therefore these orbits are also called energy levels (first energy level, second energy level and so on).

For mono electronic atoms like hydrogen atoms, the single electron initially revolves in the first shell (n=1). The electron possesses some amount of potential and kinetic energy.

When some amount of energy is given to the electron, its energy increases and it jumps to a higher energy level (i.e. n > 1) depending upon the amount of energy it absorbs. However, the excited electron is unstable and loses some of its energy by coming down to a lower level. The electron loses energy in the form of electromagnetic waves.

The reciprocal of the wavelength of the wave is given as $\dfrac{1}{\lambda }={{R}_{H}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$.

Here, ${{R}_{H}}$ is a constant for hydrogen atoms.

In a hydrogen atom, when the electron from a level that is at infinity comes down to n = 1, it is called the limit wavelength of the Lyman series.

Let the wavelength of the waves that it emits be ${{\lambda }_{1}}$.

Here, ${{n}_{1}}=1$ and ${{n}_{2}}=\infty$

Therefore, $\dfrac{1}{{{\lambda }_{1}}}={{R}_{H}}\left( \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{(\infty )}^{2}}} \right)$
$\dfrac{1}{{{\lambda }_{1}}}={{R}_{H}}\left( 1-0 \right)\Rightarrow {{\lambda }_{1}}=\dfrac{1}{{{R}_{H}}}$
It given that ${{\lambda }_{1}}=912\overset{0}{\mathop{A}}\,$
$\Rightarrow 912\overset{0}{\mathop{A}}\,=\dfrac{1}{{{R}_{H}}}$.

When the electron from a level that is at infinity comes down to n = 2, it is called the limit wavelength of the Balmer series.

Let the wavelength of the waves that it emits be ${{\lambda }_{2}}$.

Here, ${{n}_{1}}=2$ and ${{n}_{2}}=\infty$

Therefore, $\dfrac{1}{{{\lambda }_{2}}}={{R}_{H}}\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{(\infty )}^{2}}} \right)$
$\dfrac{1}{{{\lambda }_{2}}}={{R}_{H}}\left( \dfrac{1}{4}-0 \right)\Rightarrow {{\lambda }_{2}}=\dfrac{4}{{{R}_{H}}}$
But $\dfrac{1}{{{R}_{H}}}=912\overset{0}{\mathop{A}}\,$
Therefore,
${{\lambda }_{2}}=4\times 912\overset{0}{\mathop{A}}\,$.

Hence, the correct option is C.

Note: Lyman series is a spectrum of electromagnetic waves emitted by the electron when it comes down from an energy level where n > 1 to n = 1.

Lyman series is a spectrum of electromagnetic waves emitted by the electron when it comes down from an energy level where n > 2 to n = 2 .

The energy of the wave is maximum when the gap between the energy levels is maximum. Therefore, the wave with the series limit wavelength has the maximum energy.
Since the wavelength of an EM wave is inversely proportional to its energy, the limit wavelength is the minimum wavelength of the spectrum.