A nucleus at rest emits an ∝ particle. As a result
the nucleus recoils backwards, then K.E. of recoil
of the nucleus will be :
(a) greater than that of the ∝-particle
(b) less than that of the ∝-particle
(c) equal to that of the ∝-particle
(d) depends on mass number of the nucleus
Answers
Answered by
0
Answer;Mass∝Mass number or mass=xA,where A is mass number
Thorium, m
t
=234x and mass of alpha particle, m
a
=4x
Velocity of alpha particle is v
a
=1.4×10
7
m/s and that of thorium is v
t
Applying momentum conservation, m
a
v
a
+m
t
v
t
=0(rest)
so v
t
=−
m
t
m
a
v
a
, after putting values of masses we get
v
t
=−
117
2v
a
=−2.4×10
5
m/s
Magnitude of velocity of residue nucleus i.e. Thorium nuclei is found to be 2.4×10
5
m/s.
Answered by
0
Answer:
Less than that of emitted particles
Explanation:
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