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JEE Main Physics Class 12 Atoms and Nuclei Part 3 Questions
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© examsnet.com
Question : 19
Total: 69
A star has
100
%
helium composition. It starts to convert three
4
He
into one
12
C
via triple alpha process as
4
He
+
4
He
+
4
He
⟶
12
C
+
Q
. The mass of the star is
2.0
×
10
32
kg
and it generates energy at the rate of
5.808
×
10
30
W
. The rate of converting these
4
He
to
12
C
is
n
×
10
42
s
−
1
, where
n
is ______.
[Take, mass of
4
He
=
4.0026
u
, mass of
12
C
=
12
u
]
[9 Apr 2024 Shift 1]
Your Answer:
Validate
Solution:
👈: Video Solution
4
He
+
4
He
+
4
He
⟶
12
C
+
Q
power generated
=
N
t
Q
where,
N
⟶
No. of reaction/sec.
Q
=
(
3
m
He
−
m
C
)
C
2
Q
=
(
3
×
4.0026
−
12
)
(
3
×
10
8
)
2
Q
=
7.266
MeV
N
t
=
power
Q
=
5.808
×
10
30
7.266
×
10
6
×
1.6
×
10
−
19
N
t
=
5
×
10
42
rate of conversion of
4
He
into
12
C
=
15
×
10
42
Hence,
n
=
15
© examsnet.com
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