If a \(1.50\,{\rm{cm}}\)-thick piece of lead can absorb \({\rm{90}}{\rm{.0 \% }}\) of the \({\rm{\gamma }}\) rays from a radioactive source, how many centimeters of lead are needed to absorb all but \({\rm{0}}{\rm{.100 \% }}\) of the \({\rm{\gamma }}\) rays?

The spontaneous emission of radiation in the form of particles or high-energy photons as a result of a nuclear process is known as radioactivity.

The absorption of \({\rm{\gamma }}\) rays through the lead is:

\(N = {N_0}{e^{ - \mu x}}\)

Now, when the value of:

\(x = 1.50\,{\rm{cm}}\).

Then, we have the value:

\(N = \left( {10.0\% } \right){N_0}\).

So, we get:

\(\begin{align}{e^{ - \mu (1.50\;cm)}} &= \frac{{(10.0\% ){N_0}}}{{{N_0}}}\\ &= 0.100..................(1)\end{align}\)

Then, we have:

\(\mu = \frac{{\ln (0.100)}}{{(1.50\,cm)}}.............(2)\)

Now, if \({\rm{100\% }}\) of the \({\rm{\gamma }}\) rays is remaining.

We, then have the value as: \(N = 0.00100{N_0}\).

So, we get:

\(\begin{align}{e^{ - \mu x}} &= \frac{N}{{{N_0}}}\\ &= \frac{{0.00100{N_0}}}{{{N_0}}}\\ &= 0.00100..................(3)\end{align}\)

\(\begin{align}x &= \frac{{\ln (0.00100)}}{\mu }\\ &= \frac{{\ln (0.00100)}}{{\ln (0.100)}}(1.50\;\,cm)\\ &= 4.50\,\;cm........................(4)\end{align}\)

Therefore, the thickness is: \(4.50\;\,{\rm{cm}}\).