Question

What will be the longest wavelength line in Balmer series of of spectrum of H atim

Answer 1

For longest wavelength in Balmer series, n2 = 3

1/λ = R[1/n1^2 - 1/n2^2]

1/λ = R[1/2^2 - 1/3^2]
1/λ = R[1/4 - 1/9]
λ = 36/(5R)

Answer 2

For longest wavelength in Balmer series, $n_2$ = 3

Solution:

We know that,

$\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$

Where,

R = constant for Rydberg's formula

$\frac{1}{\lambda}=R\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]$

$\frac{1}{\lambda}=R\left[\frac{1}{4}-\frac{1}{9}\right]$

$\lambda=\frac{36}{(5 R)}$

$=\frac{36}{5} \times \frac{1}{R}$

$=\frac{36}{5} \times 912 \dot{\mathrm{A}}$

$=6566.4 \ \dot{\mathrm{A}}$

$\lambda=656 \ \mathrm{nm}$

It is the longest wavelength line in balmer series.

In this question, we have made use of the Rydberg’s formula, i.e.,

$\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$

λ = photon wavelength,

R = constant for Rydberg's formula = 1.0973731568539 \times 10^7 m^(-1)

Z = atomic number

$n_1$ and $n_2$ are integers where n2 > n1

In the above equation, Z = 1