Question

The intensity of gamma radiation from a given source is 1. On passing through 36 mm of lead, it is reduced to 1/8. the thickness of lead which will reduce the intensity to 1/2 will be:
A. 18 mm
B. 12 mm
C. 6 mm
D. 9 mm

Answer 1

Hint: This question is about the gamma decay. The gamma decay follows the exponential rule. For solving this question, you will need to know the values \[lo{{g}_{e}}(\dfrac{1}{8})=-2.079\] and \[lo{{g}_{e}}(\dfrac{1}{2})=-0.693\]
Formula used:
For solving this question, we will be using the formula for the intensity of the gamma radiation, i.e.,
\[I={{I}_{0}}{{e}^{-\mu d}}\]

Complete answer:
Now, let us take a look at all the given parameters first
As we discussed above,
Intensity of the gamma radiation = \[I={{I}_{0}}{{e}^{-\mu d}}\]
Where, x is the thickness of the lead
Now,
At x = 36 mm, $\dfrac{I}{{{I}_{o}}}=\dfrac{1}{8}$
Applying the given parameters in the given formula,
\[\Rightarrow \dfrac{1}{8}={{e}^{-\mu \times 36}}\]
Taking natural log on both sides.
\[\Rightarrow -\mu \times 36=\log \left( \dfrac{1}{8} \right)\]
\[\Rightarrow -\mu \times 36=-2.079\]
\[\Rightarrow ~\mu =0.0578\]
Now, If $\dfrac{I}{{{I}_{o}}}=\dfrac{1}{2}$
Again, using the formula,
\[\Rightarrow \dfrac{1}{2}={{e}^{-0.0578\times x}}\]
Where x is new thickness of the lead
Again, taking natural log on the both side
\[\Rightarrow -0.0578\times x=\log \left( \dfrac{1}{2} \right)\]
$\Rightarrow -0.0578x~=-0.693~$
\[\Rightarrow x=12~~mm\]
So, to reduce the intensity of the gamma radiation the 1/2, 12 mm of lead must be used.

So, the correct option is Option – B, 12 mm.

Note:
A gamma ray is a penetrating source of electromagnetic radiation resulting from the radioactive decay of atomic nuclei, or gamma radiation (symbol γ or.). It consists of electromagnetic waves of the shortest wavelength and hence imparts the greatest energy of photons.