Question

What is the difference in wavelengths of the
4th and 5th lines of the Balmer series in the
spectrum of atomic hydrogen :-
(1) 131 Å
(2) 520 Å
(3) 390 Å
(4) 262 Å​

Answer 1

Answer : The correct option is, (1) $131\AA$

Explanation :

The general formula for the wavelength of spectral line emitted by a hydrogen atom, when it makes a transition from $n_2$ shell to  $n_1$ shell is,

$\frac{1}{\lambda}=R_H[\frac{1}{(n_1)^2}-\frac{1}{(n_2)^2}]$

where,

$\lambda$ = wavelength

Rydberg's Constant = $R_H=1.097\times 10^7m^{-1}=911.57\AA$

Now we have to calculate the wavelength of 4th lines of Balmer series.

In this, $n_1=2$ and $n_2=6$

$\frac{1}{\lambda_4}=911.57\times [\frac{1}{(2)^2}-\frac{1}{(6)^2}]$

$\lambda_4=4102.065\AA$

Now we have to calculate the wavelength of 5th lines of Balmer series.

In this, $n_1=2$ and $n_2=7$

$\frac{1}{\lambda_5}=911.57\times [\frac{1}{(2)^2}-\frac{1}{(7)^2}]$

$\lambda_5=3970.39\AA$

Now we have to calculate the difference in wavelengths of the  4th and 5th lines of the Balmer series.

$\Delta \lambda=\lambda_5-\lambda_4$

$\Delta \lambda=4102.065\AA-3970.39\AA=131.672\AA \approx 131\AA$

Therefore, the correct option is, (1) $131\AA$

Answer 2

Answer:

Explanation:In general, the wavelength of spectral line emittted by a hydrogen atom, when it makes a transition from n2 shell to n1 shell is:

(NOTE: n2>n1 )

1λ=R[1n21−1n22]

Here, R=1.097×107m−1

Substituting value of R and solving for λ , we get:

λ=911.571n21−1n22Å

This expression gives the wavelength in Å .

For Balmer series, n1=2

When n2=6 , we get the 4th spectral line and when n2=7 , we get the 5th spectral line.

Wavelength of the 4th spectral line will be:

λ4=911.57122−162

=9×911.572

=4102.065 Å

Wavelength of the 5th spectral line will be:

λ5=911.57122−172

=911.57×19645

=3970.39 Å

Δλ=λ4−λ5

=4102.065–3970.39

=131.672 Å