Question

A monochromatic radiation of wave length $\lambda$ is incident on a hydrogen sample in ground state. The sample subsequently emits radiation of six different wave lengths, then the value of $\lambda$ is [Use $ch = 1242\text{ eV-nm}$]

• A. 80 nm
• B. 85.5 nm
• C. 97.4 nm
• D. 100.2 nm

Hint:

Number of lines $6 \rightarrow n=4$; energy required is approximately 12.75 eV.

Answer 1

The Correct Option is C

Step 1: Concept
When an electron transitions from an excited state $n$ to lower states, the number of spectral lines (wavelengths) emitted is given by $N = \frac{n(n-1)}{2}$.

Step 2: Meaning
Given $N = 6$: $6 = \frac{n(n-1)}{2} \Rightarrow n(n-1) = 12 \Rightarrow n = 4$. This means the incident radiation excited the electron from $n=1$ (ground state) to $n=4$.

Step 3: Analysis
The energy of the incident photon must be $E = E_4 - E_1$. In a hydrogen atom, $E_n = -\frac{13.6}{n^2}\text{ eV}$. $E = -0.85\text{ eV} - (-13.6\text{ eV}) = 12.75\text{ eV}$. Using the relation $E = \frac{ch}{\lambda}$: $\lambda = \frac{1242\text{ eV-nm}}{12.75\text{ eV}}$.

Step 4: Conclusion
$\lambda = 97.41... \approx 97.4\text{ nm}$. Final Answer: (C)