Question

In a radioactive material the activity at time t₁ is R₁ and at a later time t₂, it is R₂. If the decay constant of the material is \(\lambda\), then :

• A. R₁ = R₂ \(\text{e}^{-\lambda(t_1-t_2)}\)
• B. R₁= R₂ \(\text{e}^{\lambda(t_1-t_2)}\)
• C. R₁= R₂ \(\text{e}^{\bigg(\frac{t_2}{t_1}\bigg)}\)
• D. R₁ = R₂

Answer 1

The Correct Option is A

Radioactive Decay Constant: The radioactive decay of a material follows the formula: N(t) = N0 \(\times\) \(\text{e}^{(-\lambda t)}\), where N(t) is the activity at time t, N0 is the initial activity,\(\lambda\) is the decay constant, and t is time. 
If R₁ is the activity at time t₁ and R₂ is the activity at time t₂, we have: 
R₁ = N0 \(\times\) \(\text{e}^{(-\lambda t_1)}\) 
R₂ = N0 \(\times\) \(\text{e}^{(-\lambda t_2)}\) 
Dividing these equations: \(\frac{\text{R}_1}{\text{R}_2}\) = \(\frac{\text{e}^{(-\lambda t_1)}}{\text{e}^{(-\lambda t_2)}}\)= \(\text{e}^{(-\lambda(t_1-t_2))}\) 

So, the correct option is R₁ = R₂ \(\text{e}^{-\lambda(t_1-t_2)}\)