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

(a) We have,
`(mv^(2))/(r )=(Ze^(2))/(4pi epsilon_(0)r^(2))`
or, `v^(2)r=(Ze^(2))/(4 pi epsilon_(0)m)`......(i)
The quantization rule is `vr=(nh)/(2 pim)`
The radius is `r=((vr)^(2))/(v^(2)r)=(4 pi epsilon_(0)m)/(Ze^(2))`
`=(n^(2)h^(2)epsilon_(0))/(Zpi me^(2))`.....(ii)
For the given system, `Z=3` and `m=208 m_(e )`.
Thus `r_(mu)=(n^(2)h^(2)epsilon_(0))/(624pim_(e)e^(2))`
(b) From (ii), the radius of the first Bohr orbit for the hydrogen atom is
`r_(h)=(h^(2)epsilon_(0))/(pim_(e)e^(2))`
For `r_(mu)=r_(h)' (n^(2)h^(2)epsilon_(0))/(624pim_(e)e^(2))=(h^(2)epsilon_(0))/(pim_(e )e^(2))`
or, `n^(2)=624`
or, `n=25`
(c ) From (i), the kinetic energy of the atom is
`(mv^(2))/(2)=(Ze^(2))/(8 pi epsilon_(0)r)`
and the potential energy is `-(Ze^(2))/(4pi epsilon_(0)r)`
The tolal energy is `E_(n)=(Ze^(2))/(8pi epsilon_(0)r)`
Using (ii) , `E_(n)= -(Z^(2)pi me^(4))/(8 piepsilon_(0)^(2)n^(2)h^(2))= -(9xx208m_(e)^(4))/(8 epsilon_(0)^(2)n^(2)h^(2))=(1872)/(n^(2))(-(m_(e )e^(4))/(8 epsilon_(0)^(2)h^(2)))`
But `(-(m_(e)e^(4))/(8epsilon_(0)^(2)h^(2)))` is the ground state energy of hydrogen atom and hence is equal to `-13.6 eV`.
From (iii), `E_(n)= -(1872)/(n^(2))xx13.6 eV=(-25459.2eV)/(n^(2))`
Thus, `E_(1)= -25459.2 eV` and `E_(3)=(E_(1))/(9)= -2828.8eV`. The energy difference is `E_(3)-E_(1)=22630.4 eV`
The wavelength emitted is
`lambda=(hc)/(DeltaE)=(1240eV-nm)/(22630.4 eV)=55"pm"`