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Question: In a radioactive substance has half-life of h days, then its initial decay rate is given by Note tha...

In a radioactive substance has half-life of h days, then its initial decay rate is given by Note that at t = 0, M = m₀

A

m0h\frac{m₀}{h}(log 2)

B

(m₀h)(log 2)

C

-m0h\frac{m₀}{h}(log 2)

D

(-m₀h)(log 2)

Answer

-m0h(log⁡2)\frac{m_0}{h}(\log 2)

Explanation

Solution

For a radioactive substance with half-life hh days, the decay constant is

λ=ln⁡2h.\lambda = \frac{\ln 2}{h}.

The decay law is

M(t)=m0e−λt.M(t) = m_0 e^{-\lambda t}.

At t=0t = 0, the initial decay rate is given by differentiating M(t)M(t):

dMdt=−λm0e−λt.\frac{dM}{dt} = -\lambda m_0 e^{-\lambda t}.

At t=0t = 0:

dMdt∣t=0=−λm0=−m0hln⁡2.\frac{dM}{dt}\bigg|_{t=0} = -\lambda m_0 = -\frac{m_0}{h}\ln 2.

Thus, the correct initial decay rate is

−m0h(log⁡2).-\frac{m_0}{h}(\log 2).

Explanation of the solution (minimum):

  • Decay constant: λ=ln⁡2h\lambda = \frac{\ln2}{h}.
  • dMdt=−λm0\frac{dM}{dt} = -\lambda m_0 at t=0t=0 gives −m0h(ln⁡2)-\frac{m_0}{h}(\ln 2).