In a nuclear reaction 2 deuteron nuclei combine to form a helium nucleus. The energy released in $$\mathrm{MeV}$$ will be: (Given mass of deuteron $$=2.01355 \mathrm{~amu}$$. and mass of helium nucleus $$=4.0028 \mathrm{~amu}$$.

A particle at rest decays in to two particles of mass $$m_1$$ and $$m_2$$ and move with velocities $$v_1$$ and $$v_2$$. The ratio of their de Broglie wave length $$\frac{\lambda_1}{\lambda_2}$$ is:

The ground state energy of hydrogen atom is $$-13.6 \mathrm{~eV}$$. If the electron jumps from the $$3^{\text {rd }}$$ excited state to the ground state then the energy of the radiation emitted will be:

In the head-on collision of two alpha particles $$\alpha_1$$ and $$\alpha_2$$ with the gold nucleus, the closest approaches are 31.4 fermi and 94.2 fermi respectively. Then the ratio of the energy possessed by the alpha particles $$\alpha_2 / \alpha_1$$ is: