Question

The wavenumber of the first emission line in the Balmer series of H-Spectrum is:
$[R=$ Rydberg constant]

• A. $\frac{5}{36}R$
• B. $\frac{9}{400}R$
• C. $\frac{7}{6}R$
• D. $\frac{3}{4}R$

Answer 1

The correct option is A $\frac{5}{36}R$

$\overline{v}=R{Z}^2(\frac{1}{2^2}-\frac{1}{3^2})$

$=R(\frac{1}{4}-\frac{1}{9})=\frac{5R}{36}$

Hence, the correct option is $A$