A free hydrogen atom after absorbing a photon of wavelength ___ gets excited from the state n = 1 to the state n = 4. Immediately after that the electron jumps to n = m state by emitting a photon of wavelength ___.
Let the change in momentum of atom due to the absorption and the emission are Æp_ and Æp_ respectively.
If ___/___ =1/5,which of the option(s) is /are correct ?
[Use hc = 1242 eV nm; 1 nm = 10__m, h and c are PlanckÕs constant and speed of light, respectively]
A. Æp_ =1/2
Æp_
B. The ratio of kinetic energy of the electron in the state n = m to the state n = 1 is 1/4
C. ___ =418 nm
D. m = 2
Answers
Answer:
λ
e
λ
a
=
E
4
−E
1
E
4
−E
m
=
1−
16
1
m
2
1
−
16
1
=
5
1
on solving we will get
m=2
λ
e
=
13.6×3
1242×16
≈487nm
K
1
K
2
=
2
2
1
2
=
4
1
as kinetic energy is proportional to
n
2
1
Answer:
A free hydrogen atom after absorbing a photon of wavelength λ
a
gets excited from the state n=1 to the state n=4. Immediately after that the electron jumps to n = m state by emitting a photon of wavelength λ
e
. Let the change in momentum of atom due to the absorption and the emission are ΔP
a
and ΔP
e
respectively. If λ
a
/λ
e
=1/5, which of the option (s) is/are correct ?
[Use hc=1242eV nm ; 1nm=10
−9
m, h and c are Plank's constant and speed of light, respectively]
This question has multiple correct options
October 15, 2019avatar
Monalisa Itty
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ANSWER
λ
e
λ
a
=
E
4
−E
1
E
4
−E
m
=
1−
16
1
m
2
1
−
16
1
=
5
1
on solving we will get
m=2
λ
e
=
13.6×3
1242×16
≈487nm
K
1
K
2
=
2
2
1
2
=
4
1
as kinetic energy is proportional to
n
2
1