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
b. Find the energy required to the electron from `n = 3" to" n = 4`
c. Find the wavelength of radiation to remove the electron from the second bohr orbit to infinity
d. Find the kinetic energy potential energy and angular momentum of the electron in the first orbit
Find the ionination energy of above electron system in electronvalt.

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`