Radioactive element decays to form a stable nuclide, then the rate of decay of reactant `((d N)/(d t))` will vary with time `(t)` as shown in figure.

Options

A.
B.
C.
D.

Answer

C

Explanation

No. of nuclide at time t is given by
`N = N_(o) e^(- lambda t)`
Where `N_(o)` = initial nuclide thus this equation is equivalent to `y = ae^(-kx)` thus correct graph is