Question

the first emission line in the H atom spectrum in the balmer series will have wave number

Answer 1

Wave number = R [(1/n1^2) - (1/n2^2)] * Z^2

first emission line of blamer series   n1 =2 , n2 =3 and Z for hydrogen = 1

       put the value in above formula

will get option one    i e  5R/36

Answer 2

Answer: 1)  $\frac{5R_H}{36}$

Explanation:

$E=\frac{hc}{\lambda}$

$\lambda$ = Wavelength of radiation

E= energy

For ist emission line in the H atom spectrum in the balmer series will be from n= 2 to n=3.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )\times Z^2$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$ = Higher energy level = 3

$n_i$= Lower energy level = 2 (Balmer series)

Z= atomic number = 1 (for hydrogen)

Putting the values, in above equation, we get

$\frac{1}{\lambda}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\times 1$

$\frac{1}{\lambda}=\frac{5R_H}{36}$

$wavenumber=\frac{1}{\lambda}=\bar{\nu}$

$\frac{1}{\lambda}=\bar{\nu}=\frac{5R_H}{36}$

Thus wavenumber for the first emission line in the H atom spectrum in the balmer series is $\frac{5R_H}{36}$