Question

The emission spectrum of hydrogen contains a few lines. The energy ($\Delta E$) of one of the lines is $0.4578 \times 10^{-18}J$. The $n_1$ and $n_2$ belonging to this line are respectively

• A. 2, 5
• B. 2, 4
• C. 2, 3
• D. 2, 6

Hint:

{Hydrogen Spectral Series:}

• $n_1=1$: Lyman (UV)
• $n_1=2$: Balmer (Visible)
• $n_1=3$: Paschen (IR)

Calculations are faster if you memorize $1/2^2 = 0.25$, $1/3^2 \approx 0.11$, $1/4^2 \approx 0.06$, $1/5^2 = 0.04$.

Answer 1

The Correct Option is A

Step 1: Rydberg's Formula for Energy:
The energy emitted during a transition from orbit $n_2$ to $n_1$ in a hydrogen atom is given by: \[ \Delta E = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Where Rydberg constant for energy $R_H \approx 2.18 \times 10^{-18} \, J$.
Step 2: Substitute Known Values:
Given $\Delta E = 0.4578 \times 10^{-18} \, J$. \[ 0.4578 \times 10^{-18} = 2.18 \times 10^{-18} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Canceling $10^{-18}$ from both sides: \[ 0.4578 = 2.18 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] \[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{0.4578}{2.18} \approx 0.21 \]
Step 3: Analyze Options (Trial Method):
Looking at the options, $n_1$ is consistently 2 (which corresponds to the Balmer series in the visible region). Let's substitute $n_1 = 2$: \[ \frac{1}{2^2} - \frac{1}{n_2^2} = 0.21 \] \[ 0.25 - \frac{1}{n_2^2} = 0.21 \] \[ \frac{1}{n_2^2} = 0.25 - 0.21 = 0.04 \] \[ n_2^2 = \frac{1}{0.04} = \frac{100}{4} = 25 \] \[ n_2 = \sqrt{25} = 5 \]
Step 4: Final Verification:
The transition is from $n_2 = 5$ to $n_1 = 2$.