Question

Hydrogen atom is excited from ground state to another state with principal quantum number equal to 4. Then the number of spectral lines in the emission spectra will be :

A. 2
B. 3
C. 5
D. 6  
Answer :   6
Solution :
The possible number of the spectral lines is given
$$ = \frac{{n\left( {n - 1} \right)}}{2} = \frac{{4\left( {4 - 1} \right)}}{2} = 6$$

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

Practice More Releted MCQ Question on
Atoms And Nuclei


Practice More MCQ Question on Physics Section