As per Bohr model , the minimum energy (in eV) required to remove electron from the ground state of daubly joinized `Li` alon `(Z = 3)` is
A
`1.51`
B
`13.6`
C
`40.8`
D
`122.4`
Text Solution
Verified by Experts
The correct Answer is:
D
KEY CONCEPT : `E_(n) = -13.6((Z^(2)))/((n^(2))) eV` Therefore , ground state energy of ionized lithium atom `(Z = 3 , n = 1)` will be `E_(1) = (-13.6 ((3)^(2))/((1)^(2))- 122.4 eV` `:.` linozation energy of an ground state of doubly ionized lithium will be `122.4eV`.
SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise Comprehension Based Questions|2 Videos
MOVING CHARGES AND MAGNETISM
SUNIL BATRA (41 YEARS IITJEE PHYSICS)|Exercise MCQs(d )|1 Videos
Similar Questions
Explore conceptually related problems
As par Bohr model, the minimum energy (in eV ) required to remove an electron from the ground state of doubly ionized Li atom (Z = 3) is
As per Bohr model, (I) minimum energy required to remove an electron from ground state of doubly ionized Li atom (Z=3) is 122.4eV . (II) energy of transition n=3 to m=2 is less than that of m=2 to n=1 . (III) minimum energy required to remove an electron from ground state of singly-ionised He atom (z=2) is 27.2eV (IV) A Trasnsition from state n=3 to n=2 in a hydrogen atom result in U-V radiation
The minimum energy required to remove an electron is called
Minimum energy required to takeout the only one electron from ground state of He^(+) is
The energy required to remove an electron in a hydrogen atom from n = 10 state is
The ratio of the energy required to remove an electron from the first three Bohr's orbit of Hydrogen atom is
If the binding energy of the electron in a hydrogen atom is 13.6 eV , the energy required to remove the electron from the first excited state of Li^(++) is
SUNIL BATRA (41 YEARS IITJEE PHYSICS)-MODERN PHYSICS-MCQ (One Correct Answer
