A particle known as mu meson has a charge equal to that of no electron and mass `208`times the mass of the electron B moves in a circular orbit around a nucleus of charge `+3e` Take the mass of the nucleus to be infinite Assuming that the bohr's model is applicable to this system (a)drive an expression for the radius of the nth Bohr orbit (b) find the value of a for which the radius of the orbit it appropriately the same as that at the first bohr for a hydrogen atom (c) find the wavelength of the radiation emitted when the u - mean jump from the orbit to the first orbit

Explanation

(i)Electrostatic force from the nucleus acts as centripetal force for the particle . Hence , we can write the following equation :
`rArr ((208m)v^2)/r=1/(4piepsilon_0)(3exxe)/r^2`
`rArr (208m)v^2=(3e^2)/(4piepsilon_0r)`…(i)
According to Bohr.s quantisation rule , angular momentum of the meson should be integral multiple of `h//2pi`. Thus
`(208m)vr=(nh)/(2pi)`
Squaring both the sides we get the following :
`(208m)^2v^2r^2=(n^2h^2)/(4pi^2)`
`(208m)^2v^2=(n^2h^2)/(4pi^2r^2)`...(ii)
Dividing equation (ii) by equation (i) we get the following :
`((208m)^2v^2)/((208m)v^2)=(n^2h^2)/(4pi^2r^2)xx(4piepsilon_0r)/(3e^2)`
`rArr 208m=(n^2h^2epsilon_0)/(3pie^2r)`
`rArr r=(n^2h^2epsilon_0)/(3pie^2(208m))`...(iii)
(ii) Radius of the first Bohr.s orbit in Hydrogen atom can be written as follows :
`rArr r_1=(h^2epsilon_0)/(pie^2m)`....(iv)
Equating equation (3) and (4) we get the following :
`(n^2h^2epsilon_0)/(3pie^2(208m))=(h^2epsilon_0)/(pie^2m)`
`rArr n^2=3xx208`=624
`rArr` n=25
(iii)For hydrogen-like atoms we can write the following expression for wavelength :
`1/lambda=RZ^2[1/n_1^2-1/n_2^2]`
R is Rydberg constant and is given by the following expression :
`R=(me^4)/(8epsilon_0^2 ch^3)`
But in this expression instead of m we need to use 208 m, hence modified value of Rydberg constant can be written as follows :
`R.=((208m)e^4)/(8epsilon_0^2ch^3)`=208 R
And Z=3 for this atom .
On substituting we get the modified expression of wavelength as follows :
`1/lambda=(208R)(3)^2[1/n_1^2-1/n_2^2]`
Here Mu-meson makes transition from third Bohr orbit to first Bohr orbit. Hence, substituting the values we get the following :
`1/lambda=208xx1.097xx10^7xx9xx[1/1^2-1/3^2]`
`rArr 1/lambda=208xx1.097xx10^7xx9xx8/9`
`rArr lambda=1/(208xx1.097xx10^7xx8)`
`=0.05478xx10^(-9)`m = 0.05478 nm