Question

Energy required for the electron excitation in $$L{i^{ + + }}$$ from the first to the third Bohr orbit is :

A. $$36.3\,eV$$
B. $$108.8\,eV$$  
C. $$122.4\,eV$$
D. $$12.1\,eV$$
Answer :   $$108.8\,eV$$
Solution :
Energy of excitation,
$$\eqalign{ & \Delta E = 13.6{Z^2}\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right)eV \cr & \Rightarrow \Delta E = 13.6{\left( 3 \right)^2}\left( {\frac{1}{{{1^2}}} - \frac{1}{{{3^2}}}} \right) = 108.8\,eV \cr} $$

Releted MCQ Question on
Modern Physics >> Atoms And Nuclei

Releted Question 1

If elements with principal quantum number $$n > 4$$  were not allowed in nature, the number of possible elements would be

A. 60
B. 32
C. 4
D. 64
Releted Question 2

Consider the spectral line resulting from the transition $$n = 2 \to n = 1$$    in the atoms and ions given below. The shortest wavelength is produced by

A. Hydrogen atom
B. Deuterium atom
C. Singly ionized Helium
D. Doubly ionised Lithium
Releted Question 3

An energy of $$24.6\,eV$$  is required to remove one of the electrons from a neutral helium atom. The energy in $$\left( {eV} \right)$$  required to remove both the electrons from a neutral helium atom is

A. 38.2
B. 49.2
C. 51.8
D. 79.0
Releted Question 4

As per Bohr model, the minimum energy (in $$eV$$ ) required to remove an electron from the ground state of doubly ionized $$Li$$ atom $$\left( {Z = 3} \right)$$  is

A. 1.51
B. 13.6
C. 40.8
D. 122.4

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