Chapter 10, Problem 2C

To determine

To explain:

The size of Bohr orbits for a “muonic” hydrogen atom is much smaller than the corresponding Bohr orbits of an ordinary hydrogen atom.

To compare:

The energy levels of muonic hydrogen atom with the ordinary hydrogen atom.

To find:

The difference in emission spectra of a “muonic” hydrogen atom and an ordinary hydrogen atom.

Answer to Problem 2C

Bohr orbits for a “muonic” hydrogen atom is much smaller than the corresponding Bohr orbits of an ordinary hydrogen atom.

For muonic hydrogen atom the energy levels are given by$E_n{}^{\mu }=\frac{-2720}{n^2}$ eV.

The emission energy of the muonic hydrogen atom will increase by a factor of 200 as compared to the ordinary hydrogen atom

Explanation of Solution

Given:

The mass of muon is 200 times larger than the electron.

Formula used:

The radius of the Bohr orbit is given by$r=n\left(\frac{\lambda }{2\pi }\right)$, where n= 1, 2, 3,….

Calculation:

According to de Broglie’s wave hypothesis the wavelength of the orbiting electron can be given as $\lambda =\frac{h}{p}=\frac{h}{m_{e}v}\left(1.1\right)$

where p is the momentum of the electron.

Circumference of the orbit is given by$2\pi r=n\lambda \left(1.2\right)$

Substituting equation (1.1) in equation (1.2) we can write$r=n\left(\frac{\lambda }{2\pi }\right)=n\left(\frac{h}{2\pi m_{e}v}\right)\left(1.3\right)$

So, from equation (1.3) we see that radius of the Bohr orbit is inversely proportional to the mass of the electron.

Since, the mass of the muon ($m_{\mu }$) is 200 times higher than that of electron $m_e$ i.e., $m_{\mu }=200m_e\left(1.4\right)$

So, using equation (1.3) and (1.4) we can see that for muonic hydrogen atom the radius of Bohr orbit is smaller compared to ordinary hydrogen atom.

For muonic hydrogen atom the energy levels are given by$E_n{}^{\mu }=\frac{-2720}{n^2}\text{eV}\left(1.5\right)$

According to Bohr orbital theory the energy of an electron in n^th orbit is given by$E_n{}^e=\frac{-13.6}{n^2}\text{eV}\left(1.6\right)$

Where $\frac{m_e{e}^4}{8{\epsilon }_0{}^2{h}^2}=13.6\text{eV}\left(1.7\right)$

So for muonic hydrogen atom as $m_{\mu }=200m_e$, the expression for energy levels can be written as $E_n{}^{\mu }=\frac{-(13.6X200)}{n^2}=\frac{-2720}{n^2}\text{eV}\left(1.8\right)$

For ordinary Hydrogen atom the emission energy is given by$E_n{}^e=13.6\left(\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2}\right)\text{eV}$

where $n_1$ and $n_2$ are any two positive integers.

Now for muonic hydrogen atom the emission energy can be given by$E_n{}^{\mu }=2720\left(\frac{1}{n_1{}^2}-\frac{1}{n_2{}^2}\right)$ eV

Conclusion:

Bohr orbits for a ‘muonic’ hydrogen atom is much smaller than the corresponding Bohr orbits of an ordinary hydrogen atom

For muonic hydrogen atom the energy levels are given by$E_n{}^{\mu }=\frac{-2720}{n^2}$ eV.

The emission energy of the muonic hydrogen atom will increase by a factor of 200 as compared to the ordinary hydrogen atom.