Heavy metals such as lead $($Z $=82)$ are excellent x$-$ray absorbers compared to

water $($composed of low Z oxygen $($Z $=8)$ and hydrogen $($Z $=1)).$ An average

value of Z $=7\mathrm{.}42$ describes water$-$rich soft tissues in the body. For safety purposes

it is often important to completely absorb x$-$rays in a protective layer, a process

called shielding. Lead is often used to construct shields and protective garments,

such as aprons, which prevent x$-$rays from hitting the entire body during localized

procedures. Taking into account only dierences in absorption due to diering

values of Z$,$ calculate how much more strongly an atom of lead absorbs x$-$rays by

the photoelectric eect in comparison with a typical atom in soft tissue. $($Here

you are ignoring density. Since lead is denser than most materials in the body,

this will lead to an even stronger relative absorption$.)$ Ans: $1400$

$($b$)$ Consider a lead apron that contains a $0\mathrm{.}5$ mm thick layer of lead. Using the

following information, compute what fraction of $140$ keV x$-$rays incident upon the

apron will be transmitted: The mass absorption coecient of lead for $140$ keV

x$-$rays is: m $=$ $/$ $=2\mathrm{.}0$ cm$2/$g$.$ The density of lead is: $=11\mathrm{.}3$ g$/$cm$3.$Ans:

$30\%$

$($c$)$ How thick would the lead shielding in a wall of an x$-$ray experimental laboratory

have to be to reduce the intensity of $8\mathrm{.}0$ keV x$-$rays to $1\%$ of its original value?

The mass absorption coecient of lead for $8\mathrm{.}0$ keV x$-$rays is: $/$ $=232$ cm$2/$g$.$

Ans: $1\mathrm{.}8103$ cm