Question

The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4$\overset{\circ }{\mathop{A }}\,$ . Calculate the short wavelength limit for the Balmer series of the hydrogen spectrum.

Answer 1

Hint: First write the Rydberg’s formula for the wavelength of spectral lines in the hydrogen spectrum. The short-wavelength limit is given already. Consider the upper value of the principal quantum number as infinite as no value is given. Find the value of R(Rydberg’s constant) and replace it for the Balmer series to get the short wavelength limit for the Balmer series of the hydrogen spectrum.

Complete step-by-step answer:
We know that Rydberg formula for the wavelength of spectral lines in hydrogen spectrum is given by:
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{\mathop{n}_{1}^{2}}-\dfrac{1}{\mathop{n}_{2}^{2}} \right)$, (Where R is the Rydberg constant), (n is the principal quantum number)
Therefore the short wavelength limit \[{{\lambda }_{L}}\] for the Lyman series would be,
$\dfrac{1}{{{\lambda }_{L}}}=R\left( \dfrac{1}{\mathop{1}_{1}^{2}}-\dfrac{1}{\mathop{\infty }_{2}^{2}} \right)=R$
When we solve this equation we get ,
R=$\dfrac{1}{913.4}{{A }^{-1}}$ .
Now we have got R,
Therefore, the shorter wavelength limit for Balmer series would be,
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right)=\dfrac{R}{2}$
${{\lambda }_{B}}=\dfrac{4}{R}=4\times 913.4\overset{\circ }{\mathop{A }}\,$
=3653.6$\overset{\circ }{\mathop{A }}\,$

Additional Information:
The Lyman series is actually a hydrogen spectral series of transitions and resulting in ultraviolet emission of lines of the hydrogen atom as an electron descends from n ≥ 2 to n = 1 (where n is the principal quantum number).
The Balmer series of spectral emission lines of the hydrogen atom is the result of electron transitions from higher levels down to the energy level with principal quantum number 2.
The distance between two successive crests of a wave is known as wavelength.

Note: Remember the difference between the Balmer series and the Lyman series. 4/R is taken as 913.4 multiplied with 4, as R’s value was inverse. The infinity square is taken as null hence 1/infinity has no value.