Atomic weight of boron is 10.81 and it has two isotopes $$_5{B^{10}}$$ and $$_5{B^{11}}.$$ Then ratio of $$_5{B^{10}}:{\,_5}{B^{11}}$$ in nature would be
A.
$$19:81$$
B.
$$10:11$$
C.
$$15:16$$
D.
$$81:19$$
Answer :
$$19:81$$
Solution :
Let the percentage of $${B^{10}}$$ atoms be $$x,$$ then average atomic weight
$$\eqalign{
& = \frac{{10x + 11\left( {100 - x} \right)}}{{100}} = 10.81 \cr
& \Rightarrow x = 19 \cr
& \therefore \frac{{{N_B}10}}{{{N_B}11}} = \frac{{19}}{{81}} \cr} $$
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