Q 11 :

Consider the following spectral lines for atomic hydrogen:

A. First line of Paschen series
B. Second line of Balmer series
C. Third line of Paschen series
D. Fourth line of Brackett series

The correct arrangement of the above lines in ascending order of energy is:      [2026]

  • A < B < C < D

     

  • D < A < C < B

     

  • D < C < A < B

     

  • C < D < B < A

     

(2)

ΔE=13.6Z2(1n12-1n22)

Seriesn1n2A)Paschen (1st line)34B)Balmer (2nd line)24C)Paschen (3rd line)36D)Brackett (4th line)48

So correct ascending order of energy of above lines is:

D<A<C<B



Q 12 :

The energy of first (lowest) Balmer line of H atom is x J. The energy (in J) of second Balmer line of H atom is :    [2026]

  • 1.35x

     

  • x²

     

  • 2x

     

  • x1.35

     

(1)

Transition of first Balmer line

n1=2; n2=3

ΔE=x=13.6(1)2[122-132]    ..........(i)

Transition of 2nd Balmer line

n1=2; n2=4

ΔE=13.6(1)2[122-142]    ..........(ii)

Divide Eq. (ii) by Eq. (i)

ΔEx=14-11614-19

ΔEx=316536

ΔEx=2720

ΔE=1.35x



Q 13 :

The wave numbers of three spectral lines of H atom are considered. Identify the set of spectral lines belonging to Balmer series.
(R = Rydberg constant)   [2026]

  • 7R144, 3R16, 16R255

     

  • 5R36, 8R9, 15R16

     

  • 3R4, 3R16, 7R144

     

  • 5R36, 3R16, 21R100

     

(4)

Balmer series line  ν¯=RHZ2[122-1n2]

if n=3  ν¯=R(1)2[122-132]=5R36

if n=4  ν¯=3R16

if n=5  ν¯=21R100