A vessel holds 4 × 10^ 23 tritium atoms. If the half-life for the decay of tritium nuclei is 12.3 y , calculate the number of decays taking place in the subsequent 10 hours .

Correct answer: A. 2.57 × 10^ 19 The number of decays in a small time interval t is given by Δ N = λ N_0 t , since t T_ 1/2 . The decay constant λ = 2 T_ 1/2 . Given T_ 1/2 = 12.3 y = 12.3 × 365 × 24 hours = 107748 hours . Substituting the values: Δ N = (0.693)/(107748) × 4 × 10^ 23 × 10 Δ N 2.57 × 10^ 19

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 vessel holds 4 × 10^ 23 tritium atoms. If the half-life for the...

A vessel holds

4 \times 10^{23}

tritium atoms. If the half-life for the decay of tritium nuclei is

12.3 \text{ y}

, calculate the number of decays taking place in the subsequent

10 \text{ hours}

Options

A $2.57 \times 10^{19}$
B $5.14 \times 10^{19}$
C $7.15 \times 10^{15}$
D $1.54 \times 10^{21}$

Correct answer

A. $2.57 \times 10^{19}$

Step-by-step solution

The number of decays in a small time interval

t

is given by

\Delta N = \lambda N_0 t

, since

t \ll T_{1/2}

.

The decay constant

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

.

Given

T_{1/2} = 12.3 \text{ y} = 12.3 \times 365 \times 24 \text{ hours} = 107748 \text{ hours}

.

Substituting the values:

\Delta N = \dfrac{0.693}{107748} \times 4 \times 10^{23} \times 10

\Delta N \approx 2.57 \times 10^{19}

Practice The Nucleus on Quantrex Academy →

More from The Nucleus