Question

In terms of Rydberg constant $R$, the shortest wavelength in the Balmer series of hydrogen, atom spectrum will have wavelength

• A. $1/R$
• B. $4/R$
• C. $3/2R$
• D. $9/R$

Answer 1

The correct option is B $4/R$

for shortest wavelength
$n_1=2$
$n_2=\infty$
$\frac{1}{\lambda }=R(\frac{1}{2^2}-0)$
$\lambda =4/R$