The volume occupied by an atom is greater than the volume of the nucleus by factor of about
A.
$${10^{10}}$$
B.
$${10^{15}}$$
C.
$${10^{1}}$$
D.
$${10^{5}}$$
Answer :
$${10^{15}}$$
Solution :
Order of Radius of atom $$ \approx {10^{ - 10}}m$$
Order of Radius of nucleus $$ \approx {10^{ - 15}}m$$
Ratio of volume of atom to volume of nucleus $$ = \frac{{{\text{volume of atom}}}}{{{\text{volume of nucleus}}}}$$
$$\eqalign{
& = \frac{{\frac{4}{3}\pi r_1^3}}{{\frac{4}{3}\pi r_2^3}} \cr
& = {\left( {\frac{{{{10}^{ - 10}}}}{{{{10}^{ - 15}}}}} \right)^3} \cr
& = {10^{15}} \cr} $$
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