Question

The approximate wave number of the spectral line of the shortest wavelength in Balmer series of atomic hydrogen will be:
(Rydberg constant $\left(R_H\right)=109678{\text{cm}}^{-1}$)

• A. $22345.8{\text{cm}}^{-1}$
• B. $37515.6{\text{cm}}^{-1}$
• C. $15519.8{\text{cm}}^{-1}$
• D. $27419.5{\text{cm}}^{-1}$

Answer 1

The correct option is D $27419.5{\text{cm}}^{-1}$

Shortest wavelength means maximum energy.
Therefore, the electronic transition involved should be
${\text{n}}_2=\infty \to {\text{n}}_1=2$ (as it belongs to the Balmer series)
We know,
$\frac{1}{\lambda }=R{Z}^2(\frac{1}{{\text{n}}_1^2}-\frac{1}{{\text{n}}_2^2})$
Where, $\lambda =\text{Wavelength};R=\text{Rydberg constant}$
$\Rightarrow \frac{1}{{\lambda }_{\text{shortest}}}=\stackrel{-}{\nu }=109678\times 1^2\times (\frac{1}{2^2}-\frac{1}{{\infty }^2})$
$\Rightarrow \stackrel{-}{\nu }=27419.5{\text{cm}}^{-1}$