Question

What is the wavelength of the radiation emitted when the electron in a H-atom jumps from n = $\infty$ to n=2?

• A. 400 nm
• B. 420 nm
• C. 350 nm
• D. 365 nm

Answer 1

Answer: (D)

365 nm

Answer 2

The wavelength of the radiation emitted when an electron in a H-atom jumps from a higher energy level ($n_i$) to a lower energy level ($n_f$) can be calculated using the Rydberg formula:

$\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$

Where: $\lambda$ = wavelength of the emitted radiation $R_H$ = Rydberg constant ($1.097 \times 10^7 \text{ m}^{-1}$) $n_i$ = initial principal quantum number = $\infty$ $n_f$ = final principal quantum number = 2

Substitute the given values into the formula: $\frac{1}{\lambda} = 1.097 \times 10^7 \text{ m}^{-1} \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)$ $\frac{1}{\lambda} = 1.097 \times 10^7 \text{ m}^{-1} \left( \frac{1}{4} - 0 \right)$ $\frac{1}{\lambda} = 1.097 \times 10^7 \text{ m}^{-1} \times \frac{1}{4}$ $\frac{1}{\lambda} = 0.27425 \times 10^7 \text{ m}^{-1}$ $\frac{1}{\lambda} = 2.7425 \times 10^6 \text{ m}^{-1}$

Now, calculate $\lambda$: $\lambda = \frac{1}{2.7425 \times 10^6 \text{ m}^{-1}}$ $\lambda \approx 0.3646 \times 10^{-6} \text{ m}$

To convert meters to nanometers (nm), multiply by $10^9$: $\lambda \approx 0.3646 \times 10^{-6} \times 10^9 \text{ nm}$ $\lambda \approx 364.6 \text{ nm}$

This value is approximately 365 nm.