An imaginary particle has a charge equal to that of an electron and mass 100 times the mass of the electron. It moves in a circular orbit around a nucleus of charge + `4 e`. Take the mass of the nucleus to be infinite. Assuming that the Bhor model is applicable to this system. (a)Derive an expression for the radius of `n^(th)` Bhor orbit. (b) Find the wavelength of the radiation emitted when the particle jumps from fourth orbit to the second 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"`