The second line of Balmer series has wavelength \(4861 Å\). The wavelength of the first line of Balmer…
The second line of Balmer series has wavelength \(4861 Å\). The wavelength of the first line of Balmer series is
- \(1216 Å\)
- \(6563 Å\)
- \(4340 Å\)
- \(4101 Å\)
Solution
Wavelength of the second line of Balmer series, \(\lambda_2=4861 Å\)
\(\frac{1}{\lambda_2}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\)
For second line, \(n=4\)
\(\begin{aligned}
& \therefore & \frac{1}{\lambda_2} & =R\left(\frac{1}{2^2}-\frac{1}{4^2}\right) \\
\Rightarrow & & \frac{1}{4861} & =R\left(\frac{1}{4}-\frac{1}{16}\right) \\
\Rightarrow & & \frac{1}{4861} & =\frac{3 R}{16} \\
& & R & =\frac{16}{3 \times 4861} \quad \ldots (i)
\end{aligned}\)
Wavelength of first line \((n=3)\) of Balmer series is given as
\(\begin{aligned}
\frac{1}{\lambda_1} & =R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R\left(\frac{5}{36}\right) \\
& =\frac{16}{3 \times 4861} \times \frac{5}{36} \quad \text { [From Eq. (i)] } \\
\Rightarrow \lambda_1 & =\frac{3 \times 4861 \times 36}{16 \times 5}=6562.35 Å \\
\simeq 6563 Å &
\end{aligned}\)
Asked in: AP EAMCET 2020 (18 Sep Shift 2)
Practice more Atomic Physics questions on Aicharya