The wavelength of photon obtained by electron transition between two levels in h-atom and singly ionised he are lambda1 and lambda2 respectively then
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Answer:
You can use the fact that a photon emitted during the transition from n = 5 to n = 2 will carry an energy
E
equal to the difference between the energies of these two states.
Knowing this, you can relate the energy of the photon with the frequency
ν
through
E
=
h
ν
(
h
is Planck's constant.)
For any state corresponding to
n
in the hydrogen atom, you get
E
n
=
−
13.6 eV
n
2
,
where
−
13.6 eV
is the approximate ground-state energy of the hydrogen atom.
So:
E
2
=
−
13.6 eV
4
=
−
3.4 eV
=
−
5.44
⋅
10
−
19
J
E
5
=
−
13.6 eV
25
=
0.544 eV
=
−
8.7
⋅
10
−
20
J
So,
Δ
E
=
4.57
⋅
10
−
19
J
.
In
E
=
h
ν
,
ν
=
4.57
⋅
10
−
19
J
6.63
⋅
10
−
34
J
⋅
s
=
6.892
⋅
10
14
s
−
1
,
but
c
=
λ
ν
(
c
is the speed of light);
So,
λ
=
3
⋅
10
8
m/s
6.892
⋅
10
14
s
−
1
×
10
9
nm
1 m
=
435 nm
−−−−−−
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