A single electron orbits around a stationary nucleus of charge `+ Ze` where `Z` is a constant and e is the magnitude of electronic charge. It requires 47.2 eV to 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 excite the electron from `n = 3" to " n = 4`
c. Find the wavelength of radiation required 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
e. Find the ionisation energy of above electron system in electron-volt.

Answer

A, B, C, D

Explanation

a.The transition is `n_(1) = 2 rarr n_(2) = 1` by absorbing a photon of energy `47.2 eV`
`rArr Delta E = 47.2 eV`
Using the relation
`Delta E = 13.6 Z^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))eV`
`rArr 47.2 = 13.6 Z^(2) ((1)/(2^(2)) - (1)/(3^(2))) rArr Z = 5`
b.The required transition is `n_(1) = 3 rarr n_(2) = 4` by absorbing a photon of energy `Delta E`
Find `Delta E`by using the relation
`Delta E = 13.6 Z^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
`rArr Delta E= 13.6(5)^(2) ((1)/(3^(2)) - (1)/(4^(2)))`
` rArr Delta E =16.33 eV`
c.The required transition is `n_(1) = 2 rarr n_(2) = oo` by absorbing a photon of energy `Delta E`
Find `Delta E`by using the relation
`Delta E = 13.6 (5)^(2) ((1)/(2^(2)) - (1)/(oo^(2)))`
`rArr Delta E= 85 eV`.Find `lambda` of radiation corresponding to energy ` 85 eV`
`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`
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(- 340) = - 680 eV`
e. Ionization energy `= Delta E = 13.6 (5)^(2)((1)/(2^(2)) - (1)/(oo^(2)))`
`:. Delta E = 85 eV`