Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
The radius of a nucleus is given by r0A1/3, where r0 = 1.3 × 10-15 m and A is the mass number of the nucleus. The lead nucleus has A =206. The electrostatic force between two protons in this nucleus is approximately
Q. The radius of a nucleus is given by
r
0
A
1/3
, where
r
0
=
1.3
×
1
0
−
15
m
and
A
is the mass number of the nucleus. The lead nucleus has
A
=
206
. The electrostatic force between two protons in this nucleus is approximately
1892
169
KVPY
KVPY 2016
Report Error
A
1
0
2
N
0%
B
1
0
7
N
0%
C
1
0
12
N
100%
D
1
0
17
N
0%
Solution:
Taking protons at dimetrically opposite points of nucleus,
Force of electrostatic repulsion,
F
=
r
2
k
(
e
)
(
e
)
=
(
r
0
A
1/3
)
2
k
e
2
=
r
0
2
⋅
A
2/3
k
⋅
e
2
Substituting values in above equation, we get
F
=
(
13
×
1
0
−
15
)
2
(
206
)
2/3
9
×
1
0
9
×
(
1.6
×
1
0
−
19
)
2
=
0.039
×
1
0
2
N