# A particle of charge equal to that of an electron, - e, and mass 208 times 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). Assuming that the Bohr model of the atom is applicable to this system. Derive an expression for the radius of the nth Bohr orbit. Find 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. Find the wavelength of the radiation emitted when the mu-meson jumps from the third orbit of the first orbit.

Kevin Nash
10 years ago
Hello Student,
(i) Let m be the mass of electron. Then the mass of meson is 208 m. According to Bohr’s postulate, the angular momentum of mu-meson should be an integral multiple of h/2π.
∴ (208M) vr = nh/ 2π
∴ v = nh / 2π x 208 mr = nh / 416 πmr …(i)
Note : Since mu-meson is moving in a circular path, therefore, it needs centripetal force which is provided by the electrostatic force between the nucleus and mumesion.
∴ (208m)v2 / r = 1 / 4πε0 x 3e x e / r2
∴ r = 3e2 / 4πε0 x 208 mv2
Substituting the value of v from (1), we get
r = 3e2 x 416 πmr x 416 πmr / 4 πε0 x 208mn2 h2
⇒ r = n2 h2 ε0 / 624 πme2 …(i)
(ii) The radius of the first orbit of the hydrogen atom
= ε0 h2 / πme2 ….(ii)
To find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for hydrogen atom, we equate eq. (i) and (ii)
N2 h2 / 624 πme2 = ε0h2 / πme2 ⇒ n = √624 ≈ 25
(iii) 1 / λ = 208 R x Z2 [1 / n21 – 1 / n22]
⇒ 1 / λ = 208 x 1.097 x 107 x 32 [1/12 – 1/ 32]
⇒ λ = 5.478 x 10-11 m
Thanks
Kevin Nash