Question

Which one of the following electron transition between energy levels in the hydrogen atom produces the line of shortest wavelength in hydrogen spectrum?

• A. $n_2>n_1$
• B. $n_3>n_1$
• C. $n_4>n_1$
• D. $n_4>n_3$

Answer 1

Answer: (C)

$n_4>n_1$

Wavelength of transition for $H$ or $H$-like atoms is given by:

$\lambda =\frac{1}{-R{Z}^2(1/n_f^2-1/n_i^2)}$

and energy is given by:

$\Delta E=-R(1/n_f^2-1/n_i^2)$

shortest wavelength means highest transition energy

For $n_i=2,n_f=1$

$\Delta E=0.75R$

For $n_i=3,n_f=1$

$\Delta E=0.89R$

For $n_i=4,n_f=1$

$\Delta E=0.94R$

For $n_i=4,n_f=3$

$\Delta E=0.05R$

thus shortest wavelength or highest energy is for transition $4\to 1$

option C is correct