Question 529: Atoms & Nuclei

Question

The wavelength of the first line in Balmer series in the hydrogen spectrum is ' $\lambda$ '. What is the wavelength of the second line in the same series?

Accepted Answer

Wavelength in Balmer series of hydrogen spectrum is given by relation,
$$\frac{1}{\lambda}=R\left[\frac{1}{2^2}-\frac{1}{n^2}ight] \quad 	ext { where, } n=3,4,5$$
So, wavelength of first line,
$$\begin{array}{rlr}
\frac{1}{\lambda_1} & =R\left[\frac{1}{4}-\frac{1}{3^2}ight] \
& =\frac{5 R}{36} \
\Rightarrow \quad R & =\frac{36}{5 \lambda} & \left(	herefore \lambda_1=\lambda, 	ext { given) ..... (i)}ight)
\end{array}$$
Similarly wavelength of second line,
$$\begin{aligned}
& \frac{1}{\lambda_2}=R\left[\frac{1}{4}-\frac{1}{4^2}ight]=\frac{3 R}{16} \
& \Rightarrow \quad \lambda_2=\frac{16}{3 R}\quad 	ext{.... (ii)}
\end{aligned}$$
From Eqs. (i) and (ii), we get
$$\lambda_2=\frac{20}{27} \lambda 	ext {. }$$
Hence, the wavelength of the second line in the same series is $\frac{20}{27} \lambda$.

Answer

$\frac{20}{27} \lambda$

Answer

$\frac{3}{16} \lambda$

Answer

$\frac{5}{36} \lambda$

Answer

$\frac{3}{4} \lambda$

Atoms & Nuclei

The wavelength of the first line in Balmer series in the hydrogen spectrum is ' \lambda '. What is the wavelength of the second line in the same series?

Choose the correct answer: