During the nuclear explosion, one of the products is $$^{90}$$Sr with half life of 6.93 years. If 1$$\mu$$g of $$^{90}$$Sr was absorbed in the bones of a newly born baby in place of Ca, how much time, in years, is required to reduce it by 90% if it is not lost metabolically

The radioactive decay of any nuclide is governed by the exponential law

$$N \;=\; N_0\,e^{-\lambda t},$$

where $$N_0$$ is the initial number of nuclei, $$N$$ is the number left after time $$t$$, and $$\lambda$$ is the decay constant.

It is often more convenient to write this law using the half-life $$T_{1/2}$$. The relationship between the decay constant and the half-life is

$$\lambda \;=\; \frac{\ln 2}{T_{1/2}}.$$

Eliminating $$\lambda$$ from the first equation gives the form that directly involves the half-life:

$$N \;=\; N_0\Bigl(\tfrac12\Bigr)^{t/T_{1/2}}.$$

In the present problem we are told that the half-life of $$^{90}\text{Sr}$$ is $$T_{1/2}=6.93\text{ yr}$$. A newborn baby absorbs an initial mass of $$1\,\mu\text{g}$$ of this isotope. We are asked how long it will take for this quantity to drop by 90 %, that is, for only 10 % of the original amount to remain.

Mathematically, “drop by 90 %’’ means

$$\frac{N}{N_0}=0.10.$$

Substituting this fraction and the given half-life into the decay equation, we have

$$0.10 \;=\;\Bigl(\tfrac12\Bigr)^{t/6.93}.$$

To solve for $$t$$ we take the natural logarithm of both sides:

$$\ln(0.10) \;=\; \ln\!\Bigl[\Bigl(\tfrac12\Bigr)^{t/6.93}\Bigr].$$

Using the property $$\ln(a^b)=b\,\ln a$$, the right-hand side simplifies:

$$\ln(0.10) \;=\; \frac{t}{6.93}\,\ln\!\Bigl(\tfrac12\Bigr).$$

Now we isolate $$t$$:

$$t \;=\; 6.93 \times \frac{\ln(0.10)}{\ln(1/2)}.$$

Both logarithms are negative, so their ratio is positive. We calculate them explicitly:

$$\ln(0.10) = -2.302585093,\qquad \ln\!\Bigl(\tfrac12\Bigr) = -0.693147181.$$

Dividing gives

$$\frac{\ln(0.10)}{\ln(1/2)} = \frac{-2.302585093}{-0.693147181} = 3.321928094.$$

Multiplying by 6.93 yr, we find

$$t = 6.93 \times 3.321928094 = 23.02296\text{ yr}.$$

Rounding appropriately, the required time is

$$t \;\approx\; 23.03\text{ years}.$$

So, the answer is $$23.03\text{ years}$$.