Question

The intensity of gamma radiation from a given source is I. On passing through 36mm of lead, intensity is reduced to $\dfrac{I}{8}$. The thickness of lead which reduces the intensity to $\dfrac{I}{2}$ is
A 6 mm
B 9 mm
C 18 mm
D 12 mm

Answer 1

Hint: Intensity of a ray passing through ‘x’ thickness of lead is given by I=IO ${e^{ - \mu x}}$. Where IO is the intensity from the source, x is the thickness of lead.

Complete step by step answer:
Given, intensity of gamma radiation from the source, IO= I
We know that the intensity of a ray passing through ‘x’ thickness of lead is given by I=IO ${e^{ - \mu x}}$
Given that when the thickness of lead i.e. x=36 mm, the intensity i.e. I=$\dfrac{I}{8}$
Substituting the values given in the equation, we get, $\dfrac{1}{8} = I{e^{ - \mu 36}}$
Taking logarithm on both sides we get,
${\log _e}\dfrac{1}{8} = {\log _e}^{ - \mu 36}$
Using the values ${\log _e}\dfrac{1}{8} = - 2.079$ we get,
$ - 2.079 = - \mu 36$
$\implies \mu = \dfrac{{2.079}}{{36}}$
$\implies \mu = 0.0578$
Now let the thickness of lead box be x so that intensity becomes, I= $\dfrac{I}{2}$
Substituting the value of I and $\mu $ in the equation we get,
$\dfrac{I}{2} = I{e^{ - 0.0578x}}$
Then the above equation becomes, $\dfrac{1}{2} = {e^{ - 0.0578x}}$
Taking logarithm of both sides, ${\log _e}\dfrac{1}{2} = - 0.0578x$
We know that ${\log _e}\dfrac{1}{2} = - 0.693$
Then the equation becomes, -0.693=-0.0578x
$x = \dfrac{{0.693}}{{0.0578}}$
$\therefore x=12mm.$

So, the correct answer is “Option D”.

Additional Information:
The penetration power of gamma radiation is very high. They can easily pass through the body and thus pose a formidable radiation protection challenge, requiring shielding made from dense materials such as lead or concrete.

Note:
Logarithm value of common terms should be learnt by the students to solve such types of questions. Students should be familiar with some common results of logarithm. For example-${\log _e}e = 1$, ${\log _e}{e^x} = x{\log _e}e = x$ etc.