Question

A radio nuclide with decay constant $\lambda$ is being produced in a nuclear reactor at a rate q­ t per second, where $q_0$ is positive constant and t is the time. During each decay $E_0$ energy is released. The production of radio nuclide starts at time t = 0
Which differential equation correctly represents the above process

• A. $\frac{dN}{dt}+\lambda N=q_{0}t$
  
• B. $\frac{dN}{dt}-\lambda N=q_{0}t$
  
• C. $\frac{dN}{dt}{q}_{0}t=\lambda N$
  
• D. $\frac{dN}{dt}{q}_{0}t=-\lambda N$

Answer 1

The correct option is A $\frac{dN}{dt}+\lambda N=q_{0}t$


$\frac{dN}{dt}=q_{0}t-\lambda N\frac{dN}{dt}+\lambda N=q_{0}t$