Question

The half-life of radium is about 1600 years. Of 100 g of radium existing now, 25 g will remain unchanged after
A. 2400 years
B. 3200 years
C. 4800 years
D. 6400 years

Answer 1

Hint: The half-life of a given sample can be defined as the time needed for it to decay into one half of its initial amount. The law of radioactive decay can help calculate the approximate time period for a sample to decay completely by knowing the amount of sample present at the given instance.

Formula use :
The law of radioactive decay,
\[N = {N_0}{e^{ - \lambda t}}\]
Where, \[N\] - the number of atoms present at time, t
\[{N_0}\] - the number of atoms present at \[t = 0\]
\[\lambda \] - decay constant
The half-life of the given sample,
\[{T_{1/2}} = \dfrac{{\ln 2}}{\lambda } = \dfrac{{0.6931}}{\lambda }\]
Where, \[{T_{1/2}}\] - half-life of the given sample

Complete step by step solution:
To find the time period for radium when 25 g remains of the initial 100 g, we have to rearrange the above formulae.
\[N = {N_0}{e^{ - \dfrac{{\ln 2}}{{{T_{1/2}}}}t}}\]
Since exponent and natural log are inverse functions, they get cancelled out.
\[N = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{T_{1/2}}}}}} \\ \]
Given, \[{T_{1/2}} = 1600years\], \[N = 25g\] and \[{N_0} = 100g\]
To find, \[t = ?\]
By using the above formula, we get,
\[25 = 100{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{1600}}}} \\ \]
\[\Rightarrow \dfrac{{25}}{{100}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{1600}}}} \\ \]
\[\Rightarrow \dfrac{1}{4} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{1600}}}} \\ \]
\[\Rightarrow {\left( {\dfrac{1}{2}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{1600}}}}\]
Here the bases are the same, so we can equate the powers.
\[2 = \dfrac{t}{{1600}} \\ \]
\[\Rightarrow t = 2 \times {\rm{1600}} \\ \]
\[\therefore t = 3200\,years\]

Hence, the correct answer is option B.

Note: The SI unit of \[{T_{1/2}}\] is seconds. In the above steps, when the exponent and natural log cancel each other, 2 which has a negative sign before it becomes the denominator of the equation and the term \[\dfrac{t}{{{T_{1/2}}}}\] is raised to its power. If there is no negative sign then 2 will remain as a numerator.