The wavenumber of the first emission line in the Balmer seri... | Chemistry | SolveeIt | Solveeit

Today, August 18

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

$\frac{5}{36} R$

Explanation

Solution

$\bar{ v }= RZ ^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)$
$= R \left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 R }{36}$