A single electron orbits around a stationary nucless of charge `+ Ze` where `Z` is a constant and e is the magnitude of electronic charge ,if respuires 47.2 e
is excite the electron from the second bohr orbit to the third bohr orbit
a. Find the value of Z

Explanation

Since the nucless has a chaarge `+ Ze`, the atomic nucless of the ion is Z
a. The transition is `n_(!) = 2 rarr n_(2) = 3 ` by abserbing a photon of energy `47.2 eV`
`rArr Delta E = 47.2 eV`
Using the relation :
`Delta E = 13.6Z^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))eV`
`rArr 47.2 = 13.6Z^(2)((1)/(Z^(2)) - (1)/(3^2)) rArr Z = 3`
The required transition is `n_(1) = 3 rarr n_(2) = 4` by abserbing aphoton of energy of nergy `Delta E`
Find `Delta E`by using the relation
`Delta E=13.6Z^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
`rArr Delta E=13..6(5)^(2)((1)/(3^(2)) - (1)/(4^(2)))`
`Delta E = 16.53 eV`
c. The required transition is `n_(1) = 2 rarr n_(2) rarr n_(2) = 4`by abserbed aphoton of energy` Delata E `
Find `Delta E`by using the relation
`Delta E=13..6(5)^(2)((1)/(Z^(2))- (1)/(oo^(2)))`
`rArr Delta E = 8.5 eV`
Find the `lambda` of relation corresponding to energy `85eV`
`rArr lambda =- (hc)/(E ) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(85(1.6 xx 10^(-19)))`
`= 146.25 xx 10^(-10) = 146.25 Å`
d. If energy of electron be `E_(n)` then `KE = - E_(n) ` and `PE = 2E_(n)`
`E_(n) = (-13.6Z^(2))/(n^(2)) = (-13.65 xx 5^(2))/(1^(2)) = - 340 eV`
`KE = (-340 ev) = 340 ev`
`PE = 2(-34 eV) = -680 eV`
e. Ionization energy `= Delta E = 13.6(5)^(2)((1)/(2^(2)) - (1)/(oo^(2)))`
`:. Delta E = 85 eV`