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JEE Advanced 2019 Question Paper 2 with solutions for online practice
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© examsnet.com
Question : 1
Total: 54
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
Δ
p
a
and
Δ
p
e
, respectively. If
λ
a
λ
e
=
1
5
, which of the option (s) is /are correct ?
[Use
h
c
=
1242
ev
nm
;
1
nm
10
−
9
] m, h and c are Planck’s constant and speed of light, respectively]
Δ
p
a
/
Δ
p
e
=
1
2
The ratio of kinetic energy of the electron in the state n = m to the state n = 1 is
1
4
m
=
2
λ
e
=
418
nm
Validate
Solution:
1
λ
a
=
R
(
1
−
1
16
)
1
λ
e
=
R
(
1
m
2
−
1
4
2
)
λ
a
λ
e
=
(
1
m
2
−
1
4
2
)
15
16
=
1
5
1
m
2
=
1
4
2
+
3
16
1
m
2
=
1
4
⇒
m
=
2
⇒Kinetic energy
α
1
n
2
K
m
K
1
=
1
2
2
×
1
=
1
4
13.6
(
1
4
−
1
16
)
=
1242
λ
e
13.6
(
3
16
)
=
1242
λ
e
© examsnet.com
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