Question

A hydrogen atom is in ground state. Then to get six lines in emission spectrum, wavelength of incident radiation should be

• A. $800\stackrel{{}_{\circ }}{A}$
• B. $825\stackrel{{}_{\circ }}{A}$
• C. $975\stackrel{{}_{\circ }}{A}$
• D. $1025\stackrel{{}_{\circ }}{A}$

Answer 1

Answer: (C)

$975\stackrel{{}_{\circ }}{A}$

Number of possible spectral lines emitted when an electron jumps back to ground state from $n^{th}$ orbit $=\frac{n(n-1)}{2}$
Here, $\frac{n(n-1)}{2}=6\Rightarrow n=4$
Wacelength $\lambda$ from transition from $n=1$ to $n=4$ is given by, $\frac{1}{\lambda }=R(\frac{1}{1}-\frac{1}{4^2})\Rightarrow \lambda =\frac{16}{15R}=975\stackrel{{}_{\circ }}{A}$