Question

When a hydrogen atom emits a photon
of energy 12.1 eV, the orbit angular
momentum changes by
1.05*10-34-5
O 2.11-10-34 J-S
3.16*10-34 J-s
4.22*10-34
J-s​

Answer 1

Answer:

When a hydrogen atom emits a photon of energy 12.1 eV, i.e. the energy difference between two levels hence,

E

n

2

−E

n

1

=12.1

E

n

2

=12.1+E

n

1

E

n

2

=12.1+(−13.6)

E

n

2

=−1.5eV

This energy corresponds to the third orbit hence, change in orbital momentum is:

ΔL=

2π

h

(n

2

−n

1

)

ΔL=

2π

h

(3−1)=

π

h

ΔL=

3.14

6.6×10

−34

ΔL=2.11×10

−34

Js

Explanation:

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