Balmer Series MCQs
According to Bohr’s 3rd postulate, during transition of electrons from the orbits of higher energy to orbits of lower energy, it radiates the energy. The radiated energy is equivalent to the difference of energy levels between two corresponding orbits.
This theory was explained in detailed by Balmer, when he studied the energy radiated from hydrogen atom. He found the radiated energy lies in visible spectrum. Balmer series arises due to transitions of electron from different higher orbits to the second orbit (p = 2).
The wavelength is given by,
‘
Where R = Rydberg’s constant = 1.097×107 /m and n = 3, 4……
This series lies in the visible region of spectrum. The spectral lines obtained by the transition of electrons from higher orbits 2, 3, 4 and 5 are termed as Hα, Hβ, Hγ and Hδ lines.
Q.1) The lines in Balmer series the transition of electrons from 4th orbit to 2nd orbit is known as
a) Hα lines
b) Hβ lines
c) Hγ lines
d) Hδ lines
Q.2) The condition for longest wavelength in Balmer series is….
a) p= 2, n=3
b) p= 2, n=4
c) p= 2, n=5
d) p= 2, n=∞
Q.3) The condition for shortest wavelength in Balmer series is….
a) p= 2, n=3
b) p= 2, n=4
c) p= 2, n=5
d) p= 2, n=∞
Q.4) Ratio of Longest to shortest wavelength in Balmer series is…
a) 5:9
b) 9:5
c) 25:81
d) 81:25
Q.5) The wavelength of Balmer line during transition from 6th higher orbit is….
a) 7102 A0
b) 6102 A0
c) 5102 A0
d) 4102 A0
Q.6) Range of lines in Balmer series lies in……
a) infra red region
b) visible region
c) ultraviolet region
d) x-ray region
Q.7) Wavelength of Hα lines in Balmer is….
a) 3563 A0
b) 4563 A0
c) 5563 A0
d) 6563 A0
Q.8) Wavelength of Hβ lines in Balmer is…….
a) 4761 Ao
b) 4861 Ao
c) 4961 Ao
d) 5061 Ao
Q.9) Wave number for the longest wavelength in Balmer series is……
a) 0.52 × 106 /m
b) 1.52 × 106 /m
c) 2.52× 106 /m
d) 3.52× 106 /m
Answers:
Q.1) b) Hβ lines
Q.2) a) p= 2, n=3
Q.3) d) p= 2, n=∞
Q.4) b) 9:5
Q.5) d) 4102 A0
Q.6) b) visible region
Q.7) a) d) 6563 A0
Q.8) b) 4861 A0
Q.9) b) 1.52× 106 /m