Question

One mole of radium has an activity of $\dfrac{1}{{3.7}}KiloCurie$. Its decay constant will be
A) $\dfrac{1}{6} \times {10^{ - 10}}{s^{ - 1}}$
B) ${10^{ - 10}}{s^{ - 1}}$
C) ${10^{ - 11}}{s^{ - 1}}$
D) ${10^{ - 8}}{s^{ - 1}}$

Answer 1

Hint:The process of the emission of the particles from the nuclei when the nucleus is unstable is known as radioactivity. These rays or particles are emitted when the atoms get broken into parts. In this process the energy is lost by radiation. When nuclei emits charged particles, their state and configuration can change.

Complete step-By-Step solution:
Step I:
The decay constant is denoted by using the symbol lambda $(\lambda )$. This constant is used in determining the rate at which the atom decays. It is the probability of the decay per unit time. It's S.I. unit is Becquerel and denoted by Bq.
The decay constant can be written as

$\lambda = \dfrac{A}{N}$

Where
$\lambda $is the decay constant
N is the number of particles
The value of N is $6.023 \times {10^{23}}$
And A is the activity of a radioisotope
Step II:
Curie is a non S.I. unit of radioactivity.
$1Curie = 3.7 \times {10^{10}}decay/\sec = 3.7 \times {10^{10}}Bq$
Given that the mole of radium is $ = 1mole$
Converting Curie to Becquerel,

$ = \dfrac{{\dfrac{1}{{3.7 \times {{10}^{ - 3}}}} \times 3.7 \times {{10}^{10}}}}{{6.023 \times {{10}^{23}}}}$
$ = \dfrac{1}{{6.023}} \times {10^{ - 10}}{s^{ - 1}}$

Or it can also be written as$\dfrac{1}{6} \times {10^{ - 10}}{s^{ - 1}}$
Step III:
The decay constant of radium will be $\dfrac{1}{6} \times {10^{ - 10}}{s^{ - 1}}$

$ \Rightarrow $Option A is the right answer.

Note:It is important to note that the activity of a radioactive substance is the number of disintegrations that take place in a time of one second. Also the rate at which a radioactive element decays, is known as its half life. It is the time required for any isotope to decay. The charged particles are emitted and they are named as alpha, beta and the gamma rays. These charged particles are known as ions.