How much of 45 grams of pu 234 remains after 27 hours if it’s half life is 4.98 hours

1.07 g

Explanation:

Half-life of Pu-234 = 4.98 hours

Initially present = 45 g

mass remains after 27 hours = ?

Solution:

Formula

         mass remains = 1/ 2ⁿ (original mass) ……… (1)

Where “n” is the number of half lives

To find "n" for 27 hours

                          n = time passed / half-life . . . . . . . .(2)

put values in equation 2

                          n = 27 hr / 4.98 hr

                          n = 5.4

Mass after 27 hr

Put values in equation 1

          mass remains = 1/ 2ⁿ (original mass)

          mass remains = 1/ 2^5.4 (45 g)

          mass remains = 1/ 42.2 (45 g)

          mass remains =  0.0237 x 45 g

          mass remains =  1.07 g

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