A free neutron undergoes beta-decay to a proton, having a half-life of 14 minutes . Determine the decay constant.

Correct answer: B. 8.25 × 10^ -4 s^ -1 Given half-life T_ 1/2 = 14 minutes . Converting to seconds: T_ 1/2 = 14 × 60 s = 840 s The relation between decay constant λ and half-life is: λ = 2 T_ 1/2 Substituting the values: λ = (0.693)/(840) λ = 8.25 × 10^ -4 s^ -1

Concepts Of Physics MCQ Edition [Volume 2]

Exam: Concepts Of Physics MCQ Edition [Volume 2]
Subject: Physics
Chapter: The Nucleus

Concepts Of Physics MCQ Edition [Volume 2] · Physics · The Nucleus

A free neutron undergoes beta-decay to a proton, having a half-life...

A free neutron undergoes beta-decay to a proton, having a half-life of

14 \text{ minutes}

. Determine the decay constant.

Options

A $8.25 \times 10^{-2} \text{ s}^{-1}$
B $8.25 \times 10^{-4} \text{ s}^{-1}$
C $1.21 \times 10^{3} \text{ s}^{-1}$
D $4.95 \times 10^{-2} \text{ s}^{-1}$

Correct answer

B. $8.25 \times 10^{-4} \text{ s}^{-1}$

Step-by-step solution

Given half-life

T_{1/2} = 14 \text{ minutes}

.

Converting to seconds:

T_{1/2} = 14 \times 60 \text{ s} = 840 \text{ s}

The relation between decay constant

\lambda

and half-life is:

\lambda = \dfrac{\ln 2}{T_{1/2}}

Substituting the values:

\lambda = \dfrac{0.693}{840}

\lambda = 8.25 \times 10^{-4} \text{ s}^{-1}

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