Question

A radionuclide with disintegration constant $\lambda$ is produced in a reactor at a constant rate $\alpha (=2\lambda )$ nuclei per second. At t = 0, there are no nuclei present in the reactor. During each decay energy $E_0$ is released. 20% of this energy is utilized in increasing the temperature of water. Assume that there is no loss of energy through water surface. (Given $E_0=100$ times the energy required to raise the temperature of mass m of water from
$0^{\circ }C$ to ${100}^{\circ }C$).

Expression of energy released in time t is

• A. $E_0[\alpha t-\frac{\alpha }{\lambda }(1-e^{-\lambda t}\left)\right]$
  
• B. $2E_0[\alpha t-\frac{\alpha }{\lambda }(1-e^{-\lambda t}\left)\right]$
  
• C. $\frac{E_0}{2}[\alpha t-\frac{\alpha }{\lambda }(1-e^{-\lambda t}\left)\right]$
  
• D. $\frac{E_0}{4}[\alpha t-\frac{\alpha }{\lambda }(1-e^{-\lambda t}\left)\right]$

Answer 1

The correct option is A $E_0[\alpha t-\frac{\alpha }{\lambda }(1-e^{-\lambda t}\left)\right]$


Rate of formation of nuclei at time t is $\frac{dN}{dt}=\alpha -\lambda N$, Where N = number of nuclei at that time.
Integrating we get the number of nuclei left undecayed after time t to be,
$N=\frac{\alpha }{\lambda }(1-e^{-\lambda t})$
Number of nuclei disintegrated in time t,
$N_d=\alpha t-\frac{\alpha }{\lambda }(1-e^{-\lambda t})$
$\therefore$ Energy released till time t,
$E=E_0{N}_d=E_0[\alpha t-\frac{\alpha }{\lambda }(1-e^{-\lambda t}\left)\right]$