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The intensity of gamma radiation from a given source is I. On passing through 36mm of lead, intensity is reduced to I8. The thickness of lead which reduces the intensity to I2 is
A 6 mm
B 9 mm
C 18 mm
D 12 mm

Answer
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Hint: Intensity of a ray passing through ‘x’ thickness of lead is given by I=IO eμx. 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 eμx
Given that when the thickness of lead i.e. x=36 mm, the intensity i.e. I=I8
Substituting the values given in the equation, we get, 18=Ieμ36
Taking logarithm on both sides we get,
loge18=logeμ36
Using the values loge18=2.079 we get,
2.079=μ36
μ=2.07936
μ=0.0578
Now let the thickness of lead box be x so that intensity becomes, I= I2
Substituting the value of I and μ in the equation we get,
I2=Ie0.0578x
Then the above equation becomes, 12=e0.0578x
Taking logarithm of both sides, loge12=0.0578x
We know that loge12=0.693
Then the equation becomes, -0.693=-0.0578x
x=0.6930.0578
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-logee=1, logeex=xlogee=x etc.