Question 61: Atoms & Nuclei

Question

When the speed of light becomes $$\frac{2}{3}$$ of its present value, then the energy released in a given atomic explosion would

Accepted Answer

Energy released in the nuclear process is
$$E = m{c^2} - m{c^2}{\left( {{2 \over 3}} \right)^2} = \left( {1 - {4 \over 9}} \right)m{c^2} = {5 \over 9}m{c^2}$$
Hence, energy released decrease by a factor of $$\frac{5}{9}$$.

Answer

decrease by a factor $$\frac{2}{3}$$

Answer

decrease by a factor $$\frac{4}{9}$$

Answer

decrease by a factor $$\frac{5}{9}$$

Answer

decrease by a factor $$\frac{\sqrt5}{9}$$

Atoms & Nuclei

When the speed of light becomes $\frac{2}{3}$ of its present value, then the energy released in a given atomic explosion would

Choose the correct answer: