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JEE Advanced 2020 Full Test 6 Paper 2
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© examsnet.com
Question : 32
Total: 54
Which of the following option is/are correct?
Atomic orbitals are completely described as the regions where the probability of finding the electron is maximum.
The weighted average of large number of observations for measuring the radius of
1
s
orbital is greater than
52.9
p
m
(
r
2
R
2
dr represents the total probability of finding the electron between
r
and
r
+
d
r
)
.
The energy of
4
s
is always lower than
3
d
for multi electronic atom/ ion.
Energy needed to excite an electron from
n
=
2
to
n
=
4
state is
25
28
times the energy needed to excite an electron from
n
=
2
to
n
=
5
for a single electron atom
∕
ion.
Validate
Solution:
A. O. is a single
e
−
wave function
Area of the plot for
r
<
52.9
pm is smaller than the area of the plot of
r
>
52.9
p
m
.
At higher atomic number every of
3
d
<
4
s
∆
E
2
→
5
=
R
h
c
Z
2
[
1
2
2
−
1
5
2
]
=
R
h
c
Z
2
[
21
100
]
∆
E
2
→
4
=
R
h
c
Z
2
[
1
2
2
−
1
4
2
]
=
R
h
c
Z
2
×
3
16
∆
E
2
→
5
=
28
25
∆
E
2
→
4
© examsnet.com
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