Calculate the wavelength and frequency of radiation emitted when an electron in the hydrogen atom jumps from 3rd orbit to the 1st orbit.
Answers
Answer:
Start by using Rydberg formula ν~=1/λ=R(1/n2f−1/n2i) =1.10x10^7 m^-1 (1/2^2 - 1/5^2) = 2.31x10^6 this positive value means it is absorbing that energy instead of releasing it in the form of a photon.
Explanation:
Answer:
Explanation:Explanation:
The energy of the photon emitted shall equal to the difference between the electron's potential energy at the two orbits.
The
Rydberg Formula
−−−−−−−−−−−−−−− suggests that an electron that travels (or literally "falls") from energy level
n
i
to
n
f
(
n
i
>
n
f
) emits a photon of wavelength
λ
for which
1
λ
=
R
(
1
n
2
f
−
1
n
2
i
)
where the Rydberg's Constant
R
=
1.097
×
10
7
l
m
−
1
.
In this particular scenario
n
i
=
3
n
f
=
1
Such that
1
λ
=
R
(
1
n
2
f
−
1
n
2
i
)
Δ
PE
=
1.097
×
10
7
l
m
−
1
⋅
(
1
1
2
−
1
3
2
)
Δ
PE
=
9.751
⋅
10
7
l
m
−
1
λ
=
1.026
×
10
−
7
l
m
⋅
10
9
l
n
m
1
l
m
λ
=
102.6
l
n
m