Question

The luminous dials of watches are usually made by mixing a zinc sulphide phosphor with an $\alpha -$ particle emitter. The mass of radium (mass number 226, half-life 1620 years) that is needed to produce an average of $10\alpha -$ particles per second for this purpose is
(A) $2.77mg$
(B) $2.77g$
(C) $2.77 \times {10^{23}}g$
(D) $2.77 \times {10^{ - 10}}g$

Answer 1

Hint
We can find the decay constant of the radium with the help of its half-life. The rate of decay is needed to be 10. So from the first-order decay reaction, we can find the total number of atoms of radium needed. From there we can find the number of moles of radium and thereafter the total mass from the mass number 226.
In this solution, we will be using the following formula,
$\lambda = \dfrac{{\ln 2}}{{{t_{{1 \mathord{\left/ {\vphantom {1 2}} \right.} 2}}}}}$
where $\lambda$ is the decay constant and ${t_{{1 \mathord{\left/ {\vphantom {1 2}} \right.} 2}}}$ is the half-life of the radioactive atom.

Complete step by step answer
We are given a radioactive material, radium, having a half-life of 1620 years. So the half-life in terms of seconds is,
${t_{{1 \mathord{\left/ {\vphantom {1 2}} \right.} 2}}} = 1620 \times 365 \times 24 \times 3600 = 5.1 \times {10^{10}}s$
From here we can calculate the decay constant of radium from the formula,
$\lambda = \dfrac{{\ln 2}}{{{t_{{1 \mathord{\left/ {\vphantom {1 2}} \right.} 2}}}}}$
Substituting the values we get,
$\lambda = \dfrac{{\ln 2}}{{5.1 \times {{10}^{10}}s}}$
On calculating e get,
$\lambda = 1.35 \times {10^{ - 11}}{s^{ - 1}}$
From the question, we require the rate of decay of $\alpha -$ particles to be 10 particles per second. Then if $N$ is the number of particles, then
$\dfrac{{dN}}{{dt}} = 10$
Now, from the first-order decay reaction, we have,
$\dfrac{{dN}}{{dt}} = \lambda N$
So substituting the values we get,
$1.35 \times {10^{ - 11}} \times N = 10$
Therefore we can find the number of particles as,
$N = \dfrac{{10}}{{2.19 \times {{10}^{ - 8}}}} = 7.3 \times {10^{11}}$
This is the number of atoms of radium present.
So ${N_A} = 6.023 \times {10^{23}}$ is the number of atoms that are present in 1 mole.
Therefore, $7.3 \times {10^{11}}$ atoms are present in, $\dfrac{{7.3 \times {{10}^{11}}}}{{6.023 \times {{10}^{23}}}} = 1.223 \times {10^{ - 12}}$ moles.
Now in the question, it is given the atomic mass of radium is 226. So 1 mole of radium has a mass of 226g. Therefore, $1.223 \times {10^{ - 12}}$ moles of radium has a mass of $1.223 \times {10^{ - 12}} \times 226 = 2.765 \times {10^{ - 10}}g$
So the mass of radium required is $2.77 \times {10^{ - 10}}g$
The correct option is D.

Note
The half-life of a reaction is the time that is required for the reaction concentration to decrease to exactly half of the initial value. In this question, the half-life of radium is given 1620 years. For a first-order reaction, under a given set of conditions, the half-life is always constant.