Question

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

• A. $800Å$
• B. $825Å$
• C. $975Å$
• D. $1025Å$

Answer 1

Answer: (C)

Number of possible spectral lines emitted when an electron jumps back to the ground state from $n^{th}\text{orbit}=\frac{n(n-1)}{2}$ .
Here, $\frac{n(n-1)}{2}=6\Rightarrow n=4$
Wavelength $\lambda$ of an incident radiation for transition from $n=1$ to $n=4$ is given by,
$\frac{1}{\lambda }=R\left(\frac{1}{1}-\frac{1}{4^2}\right)\Rightarrow \lambda =\frac{16}{15R}=\frac{16}{15\times 1.1\times {\left(10\right)}^7}m=975Å$