Question

Average power developed in time $t$ due to the decay of the radionuclide is

• A. $(\frac{q_{0}t}{2}-\frac{q_0}{\lambda }+\frac{q_0}{{\lambda }^{2}t}-\frac{q_0}{{\lambda }^{2}t}{e}^{-\lambda t})E_0$
• B. $(\frac{q_{0}t}{2}+\frac{q_0}{\lambda }+\frac{q_0}{{\lambda }^{2}t}-\frac{q_0}{{\lambda }^{2}t}{e}^{-\lambda t})E_0$
• C. $(\frac{q_{0}t}{2}-\frac{q_0}{\lambda }+\frac{q_0}{{\lambda }^{2}t}+\frac{q_0}{{\lambda }^{2}t}{e}^{-\lambda t})E_0$
• D. $(\frac{q_{0}t}{2}+\frac{q_0}{\lambda }+\frac{q_0}{{\lambda }^{2}t}+\frac{q_0}{{\lambda }^{2}t}{e}^{-\lambda t})E_0$

Answer 1

The correct option is A $(\frac{q_{0}t}{2}-\frac{q_0}{\lambda }+\frac{q_0}{{\lambda }^{2}t}-\frac{q_0}{{\lambda }^{2}t}{e}^{-\lambda t})E_0$

The radioactive nuclide (X) is produced at rate $q_{0}t$ and decays at the rate $\lambda \left[X\right]$
Therefore, $\frac{d\left[X\right]}{dt}=q_0-\lambda \left[X\right]$
Given, Energy released per decay is $E_0$
$\Rightarrow$ Instantaneous power at any time t is $\lambda \left[X\right]E_0$
Solving, above differential equation, we get
$\left[X\right]=q_0(\frac{t}{\lambda }-\frac{1}{{\lambda }^2}(1-e^{-\lambda t}))$
Therefore, Instantaneous Power is $q_0(t-\frac{1}{\lambda }(1-e^{-\lambda t}))E_0$
Average Power:
$P_{avg}=\frac{{\int }_0^t{P}_{inst}dt}{t}$
$\Rightarrow P_{avg}=\frac{{\int }_0^t{q}_0(t-\frac{1}{\lambda }(1-e^{-\lambda t}))E_{0}dt}{t}$
Therefore, average power in time t is $q_0(\frac{t}{2}-\frac{1}{\lambda }+\frac{1}{{\lambda }^{2}t}(1-e^{-\lambda t}))E_0$