1. Advance Illustration | Atomic & Nuclear Physics | A Proton Splitting a Deutron | by Ashish Arora

1. Advance Illustration | Atomic & Nuclear Physics | A Proton Splitting a Deutron | by Ashish Arora


In this illustration we’ll analyze a proton
splitting a deutron. here we are given that when a proton of mass em p is fired on a deutron
of mass em d. having binding energy e b. splits into its constituents, we are required to
find the minimum kinetic energy of proton required for this reaction. we can also take
the mass of proton and neutron to be equal. here in this solution here we can see if a
proton having a mass em p is fired on a deutron which is having a mass em d and in this collision
when, the proton is approaching with the speed v p. we can say the maximum possible loss
in kinetic energy of proton will take place when the 2 particles colliding particles will
move together. so after collision if this is the state, before collision, and after
collision say these 2 particles of mass em p plus em d. both of these move together with
the final velocity v f if this is the state after collision then we can say, this is the
case of perfectly inelastic collision in which maximum energy loss will take place. and we
can say after collision if this deutron will transform to d star, where we consider d star
is the excited deutron. so here we can see when proton, reacts with d it transforms d
to d star, here this was at rest this was moving with some initial kinetic energy that
is, half em p v p square. and finally these 2 are moving together with the final kinetic
energy we can write half, em p plus, em d multiplied by v final square. and we can write
by conservation of momentum. here we can write em p v p was the initial momentum of this
proton that should be equal to the final momentum which is em p plus, em d. multiplied by v
final . which gives us the value of final speed with which the 2 particles have moving.
that is equal to em p v p, divided by, em p plus, em d. now in this situation, the maximum
loss in kinetic energy, which takes place here. if it is absorbed by deutron, where
this d star can be written as excited deutron nucleus. and if the absorbed energy is e b
then, this d star will split into its constituents. so, here we can write, if, d star. is, the
excited nucleus. with, absorbed energy. e b. this implies we can write initial kinetic
energy minus e b, should be equal to the final kinetic energy of, the particles and collision.
so in this situation here we can substitute the values like, here k i we can write as,
half, em p v p square. minus e b is equal to final kinetic energy is half. em p plus,
em d multiplied by v final square which is, em p v p upon, em p plus, em d, whole square.
here if we rearrange these terms, we can see. rearranging the terms here we are getting
the value of this can be written as this initial kinetic energy. so this initial kinetic energy
can be written as k i from here also i can take half em p v p square common. so this
will be k i multiplied by 1 minus, em p over em p plus, em d. is equal to e b. and further
simplifying we’ll get the value of initial kinetic energy for, this reaction to take
place in which d will absorb e b amount of energy so that it’ll further split into its
constituents,. then the minimum value of this k i can be given as, this is em p plus, em
d divided by, em d multiplied by, e b, this will be the result of this problem. so this
or more then this any kinetic energy when proton is fired onto a deutron, during reaction
it can absorb energy more then or equal to e b. so it’ll split into its constituents.

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