Q. A hydrogen atom is in the ground state. Then to get six lines in an emission spectrum, wavelength of the incident radiation should be
NTA AbhyasNTA Abhyas 2022
Solution:
Number of possible spectral lines emitted when an electron jumps back to the ground state from $n^{t h}\text{orbit}=\frac{n \left(\right. n - 1 \left.\right)}{2}$ .
Here, $\frac{n \left(\right. n - 1 \left.\right)}{2} = 6 \Rightarrow n = 4$
Wavelength $\lambda $ of an incident radiation for transition from $n = 1$ to $n = 4$ is given by,
$\frac{1}{\lambda }=R\left(\frac{1}{1} - \frac{1}{4^{2}}\right)\Rightarrow \lambda =\frac{16}{15 R}=\frac{16}{15 \times 1 . 1 \times \left(10\right)^{7}}m=975Å$
Here, $\frac{n \left(\right. n - 1 \left.\right)}{2} = 6 \Rightarrow n = 4$
Wavelength $\lambda $ of an incident radiation for transition from $n = 1$ to $n = 4$ is given by,
$\frac{1}{\lambda }=R\left(\frac{1}{1} - \frac{1}{4^{2}}\right)\Rightarrow \lambda =\frac{16}{15 R}=\frac{16}{15 \times 1 . 1 \times \left(10\right)^{7}}m=975Å$