Question 41: Atoms & Nuclei

Question

The shortest wavelengths of Paschen, Lymen and Balmer series are in the ratio

Accepted Answer

We can use the Rydberg formula for the wavelengths in the hydrogen spectrum:
$$\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}ight)$$
For each series, the shortest wavelength is when the emitted photon has the maximum energy. This is achieved when the upper energy level tends to infinity ($n_2 	o \infty$). In that case, the formula simplifies to:
$$\frac{1}{\lambda_{	ext{min}}} = R \left(\frac{1}{n_1^2} - 0ight) = \frac{R}{n_1^2} \quad \Longrightarrow \quad \lambda_{	ext{min}} = \frac{n_1^2}{R}$$
Now, letÃ¢ÂÂs see what happens for each series:
Lyman Series:
Here, $$n_1 = 1$$
Therefore,
$$\lambda_{	ext{min}} (	ext{Lyman}) = \frac{1^2}{R} = \frac{1}{R}$$
Balmer Series:
Here, $$n_1 = 2$$
Thus,
$$\lambda_{	ext{min}} (	ext{Balmer}) = \frac{2^2}{R} = \frac{4}{R}$$
Paschen Series:
Here, $$n_1 = 3$$
So,
$$\lambda_{	ext{min}} (	ext{Paschen}) = \frac{3^2}{R} = \frac{9}{R}$$
Now, comparing the three shortest wavelengths, we have:
$$\lambda_{	ext{min}} (	ext{Paschen}) : \lambda_{	ext{min}} (	ext{Lyman}) : \lambda_{	ext{min}} (	ext{Balmer}) = 9 : 1 : 4$$
Thus, the correct answer is:
Option A: $$9: 1: 4$$

Answer

$$9: 1: 4$$

Answer

$$4: 1: 9$$

Answer

$$2:1:3$$

Answer

$$3: 1: 2$$

Atoms & Nuclei

The shortest wavelengths of Paschen, Lymen and Balmer series are in the ratio

Choose the correct answer: