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 = I

1st factor of Intensity of the γ-rays = I
 = 0.4 I

2nd factor of Intensity of the γ-rays = I₂ = 12 I


To Find:

The thickness of the lead sheet = x₂ = ?



Solution:


We have the formula 

I = I₀ eµx

0.4 I₀ = I₀ eµx

or

0.4  = eµx

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

Ln(0.4) = Ln e-µx

or

-0.916 = x₁ --------------(1)


Similarly 

I = I₀ eµx

0.5 I₀ = I₀ eµx

or

0.5  = eµx

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

Ln (0.5) = Ln e-µx

or

-0.693 = x --------------(2)


Dividing Eqn (1) by (2)


-0.916-0.693 = -µx-µx


or


1.3217 = xx


or 


x = x1.3217


putting the value of x


x = 510³m1.3217


x = 3.87 x10⁻³ m 

x = 3.78 mm ------------Ans.



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