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JEE Advanced 2020 Full Test 7 Paper 2
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© examsnet.com
Question : 16
Total: 54
A hydrogen like atom (atomic number
Z
) is in a higher excited state of quantum number
n
. This excited atom can make a transition to the first excited state by successively emitting two photons of energies
10.20
e
V
and
17.00
e
V
respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies
4.25
e
V
and
5.95
e
V
respectively. Determine the value of
Z
. (lonisation energy of hydrogen atom is
13.6
e
V
).
Your Answer:
Validate
Solution:
We have
E
=
13.6
Z
2
(
1
n
1
2
−
1
n
2
2
)
(
in
e
V
)
10.2
+
17
=
13.6
Z
2
(
1
2
2
−
1
n
2
)
.
.
.
(i)
and
4.25
+
5.95
=
13.6
Z
2
(
1
3
2
−
1
n
2
)
from (i) & (ii)
n = 6 and Z = 3
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