Two radioactive substances $A$ and $B$ of mass numbers $200$ and $212$ respectively, show spontaneous $\alpha$-decay with the same $Q$ value of $1 \text{ MeV}$. The ratio of energies of $\alpha$-rays produced by $A$ and $B$ is . . . . . . .

Assertion : The ionising power of $\beta$-particle is less compared to $\alpha$-particles but their penetrating power is more.
Reason : The mass of $\beta$-particle is less than the mass of $\alpha$-particles.

The correct order of ionizing capacity of $\alpha, \beta,$ and $\gamma$ rays is:

$A$ radioactive source is placed in a uniform electric field. If $\alpha, \beta$,and $\gamma$ particles are emitted,then $\alpha, \beta, \gamma$ particles correspond to which paths respectively?

The $Q$-value of the decay ${ }_{11}^{22} Na \rightarrow{ }_{10}^{22} Ne + e ^{+}+v$ is

$A$ radio nuclide $A_1$ with decay constant $\lambda_1$ transforms into a radio nuclide $A_2$ with decay constant $\lambda_2$. If at the initial moment the preparation contained only the radio nuclide $A_1$,then the time interval after which the activity of the radio nuclide $A_2$ reaches its maximum value is: