Question

Calculate the wavenumber of the photon emitted during transition from the orbit of $n=2$ to $n=1$ in hydrogen atom. [ $R_{H}=109677\text{ cm}^{-1}$]

• A. $27419.3\text{ cm}^{-1}$
• B. $109677.0\text{ cm}^{-1}$
• C. $12064.5\text{ cm}^{-1}$
• D. $82257.8\text{ cm}^{-1}$

Hint:

Logic Tip: The fraction $\frac{3}{4}$ represents $75%$. You can quickly estimate $75%$ of $110,000$, which is $82,500$. Only option D ($82257.8$) is in this approximate range!

Answer 1

The Correct Option is D

Concept:
The wavenumber ($\bar{\nu}$) of a photon emitted during an electronic transition in a hydrogen-like atom is given by the Rydberg formula: $$\bar{\nu} = R_H \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ where $R_H$ is the Rydberg constant, $Z$ is the atomic number, $n_1$ is the lower energy level, and $n_2$ is the higher energy level.
Step 1: Identify the values for the given transition.
Atom: Hydrogen $\implies Z = 1$ Transition is from $n = 2$ to $n = 1$. Lower energy level, $n_1 = 1$ Higher energy level, $n_2 = 2$ Rydberg constant, $R_H = 109677\text{ cm}^{-1}$
Step 2: Substitute values into the Rydberg formula.
$$\bar{\nu} = 109677 \times (1)^2 \times \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$$ $$\bar{\nu} = 109677 \times \left( \frac{1}{1} - \frac{1}{4} \right)$$ $$\bar{\nu} = 109677 \times \left( 1 - 0.25 \right)$$ $$\bar{\nu} = 109677 \times \frac{3}{4}$$
Step 3: Perform the arithmetic calculation.
First, divide by 4: $$\frac{109677}{4} = 27419.25$$ Next, multiply by 3: $$\bar{\nu} = 27419.25 \times 3$$ $$\bar{\nu} = 82257.75\text{ cm}^{-1}$$ Rounding to one decimal place matches $82257.8\text{ cm}^{-1}$.