A particle known as mu mean 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 nucles to be infinite Assuming that the bohr's model is applicable to this system (a)drive an eqression for the radius of the nth Bohr orbit (b) find the value of a for which the redius of the orbit it approninately 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))/(4 pi epsilon_(0)r^(2))`
`or, `u^(2)r = (Ze^(2))/(4 pi epsilon_(0)m)`…(i)
The quantiation rule is `ur^(2) = (pi h)/(2 pi m)`
The radius is `r= ((ur)^(2))/(u^(2)r) = (n^(2)h^(2))/(4 pi^(2)m^(2)) (4 pi epsilon_(0)m)/(Ze^(2)`
`= (n^(2)h^(2)epsilon_(0))/(2 pi me^(2))`...(ii)
For the given system `Z = 3 and m = 208m_(2)`
thus `r_(mu) = (n^(2)h6(2) epsilon_(0))/(624 pi m_(e) e6(2))`
(b)from (iii) The radius of the first bohr orbit for the hydrogen atom is
`r_(h) = (h^(2)epsilon_(0))/(pi m_(e)e^(2))`
For `r_(mu) = r_(h+)`
n^(2)h^(2)epsilon_(0))/(624 pi m_(e) e6(2)) = (h^(2)epsilon_(0))/(pi m_(e) e6(2))`
`or, `n^(2) = 624`
`n = 25`
`(c)From the kinetic energy of the atom is
`mv^(2))/(2) = (Ze^(2))/(8 pi epsilon_(0)r)`
The tatal energy is `E_(0)= (Ze^(2))/(8 pi epsilon_(0)r)`
Using (ii)
`E_(0)= -(Z^(2) pi me6(2))/(8 pi epsilon_(0)^(2) pi6(2)h^(2)) = -(9 xx 208m_(e)e^(4))/(8 epsilon_(0)^(2) n6(2)h6(2))`
`= (1872)/(n^(2)) (-(m_(e)e^(4)/(8 epsilon_(0)^(2)h^(2)))`...(iii)
But` (- (m_(e)e6(1))/(8 epsilon_(0)^(2)h^(2)))` is the ground state energy of hydrogen atom and hence is equal to `- 13.6eV`
From (iii) `E_(0)= -(1872)/(n^(2)) xx 13.6eV = (-25459.2eV)/(n^(2))`
Thus `E_(1) = -25459.2 eVand E_(2) = (E_(1)/(9) = -2828.8eV`The energy difference is `E_(2) - E_(1) = 22630.4eV`
The wavelength emitted is
`lambda = (hc)/(Delta E)`
= (1242eV nm)/(22630.4eV) = 55pm`