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Classical Mechanics Level 4

The radius of the orbit of an electron in a Hydrogen-like atom is $4.5a_0$ , where $a_0$ is the Bohr's radius. It's orbital angular momentum is $\frac{3h}{2\pi}$ . It is given that $h$ is Planck's constant and $R$ is Rydberg constant. The possible wavelengths when the atom de-excites is/are:-

1) $\frac{9}{32R}$

2) $\frac{9}{16R}$

3) $\frac{9}{5R}$

4) $\frac{4}{3R}$

NOTE: To submit the answer: If the answer is 1,4 enter answer as 14. If the answer is 2,3,4 enter answer as 234, means concatenate your options in the increasing order.

The answer is 13.

1 solution

Aryaman Maithani
Jun 22, 2018

According to Bohr's model, the angular momentum of an electron is quantised and it given by $\dfrac{nh}{2\pi}$ where $n$ is the principal quantum number. From the value of momentum given, it is clear that $n=3$ .

The radius of the orbit of an electron is given by $\dfrac{n^2}{Z} a_0$ where $n$ is the same as above and $Z$ is the atomic number of the atom. $\therefore Z = 2$ here.

When an electron de-excites it releases energy in the form of a photon whose wavelength is related to the principal quantum numbers as:

$\dfrac{1}{\lambda} = RZ^2\Bigg(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\Bigg)$

Where $n_f$ and $n_i$ are the final and initial quantum numbers respectively.

Where an electron from $n=3$ de-excited, the following transitions are possible:

$3\to2$
$2\to1$
$3\to1$

Subsituting the value of $Z=2$ and checking the transitions, it can be seen that options $1$ and $3$ are correct.

Mission JEE!

The answer is 13.

1 solution

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