Question

Monochromatic radiation of wave-length $\lambda$ is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiations of six different wavelengths. Find the wavelengths $\lambda$.

• A. $975A$
• B. $1218A$
• C. $2248A$
• D. $4316A$

Answer 1

The correct option is A $975A$

Equating $6$ with $n(n-1)/2$, we get $n=4$

it means after absorbing the photon the hydrogen had gotten the state $n=4$ subsequently after $de-excitation$

it emitted $6$ spectral lines.

Now energy of state $n=4$ is $E_4=-13.6/4^2=-0.85eV$

Where$-13.6eV=E_1$

So the energy difference=$13.6-0.85=12.75eV$

If wavelength be $\lambda$ then $\lambda =\frac{hc}{E\left(eV\right)}nm=\frac{1242.3}{12.75}nm=97.43nm=974.3A$

Option A is correct.