AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

The energy of photons from state 3 to state 2 is given by,
ΔE=13.6(

1

4

−

1

9

) =

13.6×5

36

=1.89eV
The energy of photons from state 4 to state 2 is given by,
ΔE=13.6(

1

4

−

1

16

) =

13.6×12

64

=2.55eV
The energy of photons from infinite to state 2 is given by,
ΔE=13.6(

1

4

−

1

∞

) =

13.6

4

=3.4eV
The ratio of kinetic energy of emitted photons is,
r=

(2.55−1.89)eV

(3.4−1.89)eV

≈

7

16