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 = `\1/2` I₀


To Find:

The thickness of the lead sheet = x₂ = ?



Solution:


We have the formula 

I₁ = I₀ `\e^{µx₁}`

0.4 I₀ = I₀ `\e^{µx₁}`

or

0.4  = `\e^{µx₁}`

taking the natural logarithm (ln) on both sides we get 

Ln(0.4) = Ln `\e^{-µx₁}`

or

-0.916 = -µx₁ --------------(1)


Similarly 

I₂ = I₀ `\e^{µx₂}`

0.5 I₀ = I₀ `\e^{µx₂}`

or

0.5  = `\e^{µx₂}`

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)


`\frac {-0.916}{-0.693}` = `\frac {-µx₁}{-µx₂}`


or


1.3217 = `\frac {x₁}{x₂}`


or 


x₂ = `\frac {x₁}{1.3217}`


putting the value of x₁


x₂ = `\frac {5 ｘ10⁻³ m}{1.3217}`


x₂ = 3.87 ｘ10⁻³ m 

x₂ = 3.78 mm ------------Ans.



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