Question 42: Atoms & Nuclei

Question

In a hydrogen atom, if electron is replaced by a particle which is 40 times heavier but has the same charge, then, the ratio of the radius of the first excited state of a normal hydrogen atom to the ground state of the above atom is

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

40 : 1

Answer

1 : 160

Answer

1 : 40

Answer

160 : 1

Atoms & Nuclei

In a hydrogen atom, if electron is replaced by a particle which is 40 times heavier but has the same charge, then, the ratio of the radius of the first excited state of a normal hydrogen atom to the ground state of the above atom is

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