Question

A $\mu -$-meson particle of charge equal to that of an electron, $- e$ and mass 208 times the mass of the electron moves in a circular orbit around the nucleus of charge $+ 4e$. Assuming the Bohr’s model of the atom to be applicable to this system, the inverse of wavelength of photons emitted when an electron jumps in the 4th orbit of the atom from infinity in terms of Rydberg constant is given by $\alpha {R_H}$. Find the value of $\alpha /26$.

Answer 1

Use the mass-energy relation to determine the energy difference between the two orbits for meson. Since the meson has 208 times the mass of the electron, the energy difference will also be 208 times that of the electron. Use a formula for wavenumber from Bohr’s model to determine wave number for meson.

Formula used:
$\dfrac{1}{\lambda } = {R_H}{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Here, ${R_H}$ is Rydberg’s constant and $Z$ is atomic number.

Complete step by step answer: When an electron jumps from higher orbit ${n_2}$ to lower orbit ${n_1}$, the difference in the energy of the orbit is given by Bohr’s atomic model as,
$\Delta E = {R_H}{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$ …… (1)

Here, ${R_H}$ is Rydberg’s constant and $Z$ is atomic number.

We know that the difference in the energy is given as,
$\Delta E = \dfrac{{hc}}{\lambda }$

Therefore, the energy difference is inversely proportional to the wavelength of the electron. Therefore, the equation (1) is written as,
$\dfrac{1}{\lambda } = {R_H}{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$

The term $\dfrac{1}{\lambda }$ is known as wave number.

According to the mass-energy relation, the difference in the energy is,
$\Delta E = m{c^2}$

Here, m is the mass of the particle and c is the speed of light.

We have given that the mass is 208 times the mass of the electron. Therefore, the energy difference will also be 208 times that of the electron. Therefore, we can write,
$\dfrac{1}{\lambda } = 208{R_H}{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$

Substitute 4 for Z, 4 for ${n_1}$ and $\infty$ for ${n_2}$ in the above equation.
$\dfrac{1}{\lambda } = 208{R_H}{\left( 4 \right)^2}\left( {\dfrac{1}{{{4^2}}} - \dfrac{1}{{{\infty ^2}}}} \right)$
$\Rightarrow \dfrac{1}{\lambda } = 208{R_H}$

We have given that the wave-number for electrons is $\alpha {R_H}$. Since the given meson also follows Bohr’s model, we can write,
$\alpha {R_H} = 208{R_H}$
$\Rightarrow \alpha = 208$

Therefore,
$\dfrac{\alpha }{{26}} = \dfrac{{208}}{{26}} = 8$

Therefore, the value of $\dfrac{\alpha }{{26}}$ is 8.

Note: Rydberg constant ${R_H}$ consists of all the constants including mass of the electron m, charge e, speed of light c and Planck’s constant h. therefore, do not consider any other constants other than Z in the above formula. The mass-energy relation implies the energy possessed by the rest particle.