A single electron orbit around a stationary nucleus of charge `+ Ze` where Z is a constant and e is the magnitude of the electronic charge. It requires`47.2 eV` to excite the electron from the second bohr orbit to the third bohr orbit. Find
(i) The value of Z
(ii) The energy required by nucleus to excite the electron from the third to the fourth bohr orbit
(iii) The wavelength of the electronmagnetic radiation required to remove the electron from the first bohr orbit to inlinity
(iv) The energy potential energy potential energy and the angular momentum of the electron in the first bohr orbit
(v) The radius of the first bohr orbit (The ionization energy of hydrogen atom ` = 13.6 eV ` bohr radius `= 5.3 xx 10^(-11) matre` velocity of light `= 3 xx 10^(8) m//sec` planks 's constant ` = 6.6 xx 10^(-34)` jules - sec )

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`