A sheet of lead 5.0 mm thick reduces the intensity of a beam of γ-rays by a factor of 0.4. Find the half-value thickness of the lead sheet which will reduce the intensity to half of its initial value. (Ans : 3 .79 mm )
Given Data:
Thickness of the lead sheet = x₁ = 5 mm = 5 x10⁻³ m
Let the initial value of Intensity of the γ-rays = I₀
1st factor of Intensity of the γ-rays = I₁ = 0.4 I₀
2nd factor of Intensity of the γ-rays = I₂ = 12 I₀
To Find:
The thickness of the lead sheet = x₂ = ?
Solution:
We have the formula
I₁ = I₀ eµx₁
0.4 I₀ = I₀ eµx₁
or
0.4 = eµx₁
taking the natural logarithm (ln) on both sides we get
Ln(0.4) = Ln e-µx₁
or
-0.916 = -µx₁ --------------(1)
Similarly
I₂ = I₀ eµx₂
0.5 I₀ = I₀ eµx₂
or
0.5 = eµx₂
taking the natural logarithm (ln) on both sides we get
Ln (0.5) = Ln e-µx₂
or
-0.693 = -µx₂ --------------(2)
Dividing Eqn (1) by (2)
or
1.3217 = x₁x₂
or
x₂ = x₁1.3217
putting the value of x₁
x₂ = 5x10⁻³m1.3217
x₂ = 3.87 x10⁻³ m
x₂ = 3.78 mm ------------Ans.
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