a Substitute the table's value for aluminum and I(x)=0.3I_0 and solve for x.
B
b Substitute the table's value for copper and I(x)=0.3I_0 and solve for x.
C
c Substitute the table's value for lead and I(x)=0.3I_0 and solve for x.
D
d See solution.
A
a x≈ 2.80
B
b x≈ 0.38
C
c x≈ 0.03
D
d See solution.
Practice makes perfect
a Like the hint says, we want to solve for x when I(x)= 0.3I_0. Also, since we are investigating aluminum, we should substitute μ=0.43 in the formula. Then, we should isolate the term containing the variable.
Now that the term with the variable as an exponent is isolated, we should notice that the base is e. This means that, to get to the exponent, we should take the natural logarithm of both sides.
Now that the term with the variable as an exponent is isolated, we should notice that the base is e. This means that, to get to the exponent, we should take the natural logarithm of both sides.
Now that the term with the variable as an exponent is isolated, we should notice that the base is e. This means that, to get to the exponent, we should take the natural logarithm of both sides.
The thickness of the lead shielding should be 0.03 cm thick to reduce the intensity of the X-rays to 30 %.
d From previous parts, we know that the following thicknesses of aluminum, copper, and lead, are required to reduce the intensity of X-rays to 30 % of their initial intensity.
Aluminum:& 2.80 cm
Copper:& 0.38 cm
Lead:& 0.03 cm
As we can see, when lead is used, the apron does not need to be very thick to reduce the harmful radiation. Therefore, one might want to use lead-aprons instead of, for example, aprons made from aluminum.
