Question 117: Atoms & Nuclei

Question

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :

Accepted Answer

The wavelength of light emitted when an electron transitions between energy levels in a hydrogen atom is given by the Rydberg formula:
$$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$
where:
$$\lambda$$ is the wavelength of the emitted light,
$$R$$ is the Rydberg constant,
$$n_1$$ is the lower energy level, and
$$n_2$$ is the higher energy level.
The Balmer series corresponds to electron transitions where the lower energy level, $$n_1$$, is 2, and the Lyman series corresponds to transitions where $$n_1$$ is 1. The shortest wavelength in each series occurs for transitions from the highest possible energy level ($$n_2 = \infty$$) to the specified lower level ($$n_1$$).
For the Balmer series (shortest wavelength)
:
$$n_1 = 2, n_2 = \infty$$,
Putting these values into the Rydberg formula gives:
$$\frac{1}{\lambda_{B}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)$$
$$\frac{1}{\lambda_{B}} = R \left( \frac{1}{4} \right)$$
$$\lambda_{B} = \frac{1}{R \cdot \frac{1}{4}} = \frac{4}{R}$$
For the Lyman series (shortest wavelength)
:
$$n_1 = 1, n_2 = \infty$$,
Putting these values into the Rydberg formula gives:
$$\frac{1}{\lambda_{L}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)$$
$$\frac{1}{\lambda_{L}} = R \cdot 1$$
$$\lambda_{L} = \frac{1}{R}$$
The ratio of the shortest wavelength of Balmer series ($$\lambda_{B}$$) to the shortest wavelength of Lyman series ($$\lambda_{L}$$) is:
$$\frac{\lambda_{B}}{\lambda_{L}} = \frac{\frac{4}{R}}{\frac{1}{R}} = \frac{4}{1} = 4:1$$
Therefore, the correct answer is
Option D: $$4: 1$$
.

Answer

$$1: 2$$

Answer

$$1: 4$$

Answer

$$2: 1$$

Answer

$$4: 1$$

Atoms & Nuclei

The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :

Choose the correct answer: