The rate of disintegration of a radioactive sample can be increased by:

$A$ 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:

$A$ nucleus with $Z=92$ emits the following in a sequence: $\alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \alpha, \alpha, \alpha, \beta^{-}, \beta^{-}, \alpha, \beta^{+}, \beta^{+}$ and $\alpha$. The atomic number of the resulting nucleus is

Consider the decay of a free neutron at rest: $n \rightarrow p + e^-$. Show that the two-body decay of this type must necessarily give an electron of fixed energy and,therefore,cannot account for the observed continuous energy distribution in the $\beta$-decay of a neutron or a nucleus.

$A$ radioactive substance emits

The ${ }_{92}^{238} U$ atom disintegrates to ${ }_{84}^{214} Po$ with a half-life of $4.5 \times 10^9$ years by emitting $6$ $\alpha$-particles and $n$ electrons. Here,$n$ is: