When an $$\alpha $$-particle of mass $$m$$ moving with velocity $$v$$ bombards on a heavy nucleus of charge $$Ze,$$ its distance of closest approach from the nucleus depends on $$m$$ as
A.
$$\frac{1}{{\sqrt m }}$$
B.
$$\frac{1}{{{m^2}}}$$
C.
$$m$$
D.
$$\frac{1}{m}$$
Answer :
$$\frac{1}{m}$$
Solution :
When an $$\alpha $$-particle of mass $$m$$ moving with velocity $$v$$ bombards on a heavy nucleus of charge $$Ze,$$ then there will be no loss of energy as in this case, initial kinetic energy of $$\alpha $$-particle = potential energy of $$\alpha $$-particle at closest approach.
$$\eqalign{
& \Rightarrow \frac{1}{2}m{v^2} = \frac{{2Z{e^2}}}{{4\pi {\varepsilon _0}{r_0}}} \cr
& \Rightarrow \boxed{{r_0} \propto \frac{1}{m}} \cr} $$
This is the required distance of closest approach to $$\alpha $$-particle from the nucleus.
Releted MCQ Question on Modern Physics >> Atoms or Nuclear Fission and Fusion
In the nuclear fusion reaction
$$_1^2H + _1^3H \to _2^4He + n$$
given that the repulsive potential energy between the two
nuclei is $$ \sim 7.7 \times {10^{ - 14}}J,$$ the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s Constant $$k = 1.38 \times {10^{ - 23}}J/K$$ ]
The binding energy per nucleon of deuteron $$\left( {_1^2H} \right)$$ and helium nucleus $$\left( {_2^4He} \right)$$ is $$1.1\,MeV$$ and $$7\,MeV$$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is