Question

Calculate the wave number of the highest energy transition in the Balmer series of the hydrogen atom:

$[R_H=109678c{m}^{-1}]$

• A. $27419.5c{m}^{-1}$
• B. $2741.95c{m}^{-1}$
• C. $274.195c{m}^{-1}$
• D. $274195.1c{m}^{-1}$

Answer 1

The correct option is A $27419.5c{m}^{-1}$

The highest energy transition in the Balmer series of Hydrogen atom corresponds to $\infty \to 2$ transition.
$\overline{\nu }=109678c{m}^{-1}\times [\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\overline{\nu }=109678c{m}^{-1}\times [\frac{1}{2^2}-\frac{1}{{\infty }^2}]$

$\overline{\nu }=109678c{m}^{-1}\times 0.25$

$\overline{\nu }=27419.5c{m}^{-1}$

The wave number of the highest energy transition in the Balmer series of hydrogen atom is $27419.5c{m}^{-1}$.

Hence, option A is correct.