Question

A formula analogous to the Rydberg formula applies to the series of spectral lines which arise from transitions from higher energy level to lower energy level of hydrogen atom.
A muonic hydrogen atom is like a hydrogen atom in which electrons are replaced by heavier particles, the muon. The mass of the muon is 207 times the mass of the electron, while the charge remains the same as that of the electron. Rydberg formula for hydrogen atom is:
$\dfrac{1}{\lambda } = {R_H}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]$ (${R_H}$= 109678 $c{m^{ - 1}}$)
Radius of first bohr orbit of muonic hydrogen atom is :
a.) $\dfrac{{0.259}}{{207}}{A^0}$
b.) $\dfrac{{0.529}}{{207}}{A^0}$
c.) $0.529 \times 207{A^0}$
d.) $0.259 \times 207{A^0}$

Answer 1

Hint: Radius of an atom is inversely proportional to the mass of its particles. A muon is heavier than an electron. Thus, its radius will consequently decrease because the mass is large. Thus, it will have a value of radius of an electron divided by 207.

Complete step by step answer:
We have been given in question that the mass of muon is 207 times the mass of an electron.
Further, we know that radius is inversely proportional to mass. This is because as the mass of an atom increases. This will lead to increase in the number of electrons, protons etc. Thus, the nuclear charge will increase and it will attract the outer electrons with more force. So, this will result in a decrease in the radius of the atom. This can be written as –
$r = \dfrac{1}{m}$
Where r = radius of an atom and
m = mass

For finding radius of muonic by comparison with electronic, we can find as -
$\dfrac{{{r_{muonic}}}}{{{r_{electronic}}}} = \dfrac{{{m_{electron}}}}{{{m_{muon}}}}$
Here, we are given with ${r_{electronic}}$ = 0.529 ${A^0}$
Further mass of muonic is 207 times the mass of electrons.
Thus, putting the values in formula, we have
$\begin{gathered}
  \dfrac{{{r_{muonic}}}}{{0.529{\text{ }}{{\text{A}}^0}}} = \dfrac{1}{{207}} \\
  {r_{muonic}} = \dfrac{{0.529{\text{ }}{{\text{A}}^0}}}{{207}} \\
\end{gathered} $
So, the correct answer is “Option B”.

Note: We know the radius of first bohr orbit = 0.529 ${A^0}$
The radius of nth Bohr orbit can be found as –
 $^{{r_n} = \dfrac{{{n^2}{h^2}}}{{4{\Pi ^2}mZ{e^2}}}}$
where ${r_n}$ is the radius of nth orbit
n is number of orbit
m is the mass
h is planck’s constant
Try to complete the answer step by step. Always write units along with numerical values. It helps during conversions.