Question

A particle of charge equal to that of an electron, -e and mass 208 times of the mass of the electron (called a mu-meson) moves in a circular orbit around a nucleus of charge +3e. (Take the mass of the nucleus to be infinite). Assume that the Bohr model of the atom is applicable to this system. If the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom, then the value of $\frac{n}{4}$ rounded off to nearest integer is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is C

6

If we assume that mass of nucleus >> mass of mu-meson, then nucleus will be assumed to be at rest, only mu-meson is revolving around it.

In $n^{th}$ orbit, the necessary centripetal force to the mu-meson will be provided by the electrostatic force of attraction between the nucleus and the mu-meson. Hence,

$\frac{m{v}^2}{r}=\frac{1}{4\pi {ϵ}_0}\frac{\left(Ze\right)\left(e\right)}{r^2}$ …(1)

Further, it is given that Bohr model is applicable to the system. Hence,

Angular momentum in $n^{th}$ orbit $=\frac{nh}{2\pi }$

or $mvr=n\frac{h}{2\pi }$

We have two unknowns v and r (in $n^{th}$ orbit). After solving these two equations, we get

$r=\frac{n^2{h}^2{ϵ}_0}{Z\pi m{e}^2}$

Substituting Z = 3 and m = 208 m, we get $r=\frac{n^2{h}^2{ϵ}_0}{624\pi m_e{e}^2}$

The radius of the first Bohr orbit for the hydrogen atom is: $\frac{h^2{ϵ}_0}{624\pi m_e{e}^2}$

Equating this with the radius calculated in part (i), we get or $n^2\approx 624$ or $n\approx 25$