The ceramic glaze on a red-orange Fiestaware plate is U2O3 and contains 50.1 grams of 238U, but very little 235U. (a) What is the activity of the plate (in Ci)

The correct answer is 1.68 × 10⁻⁵ Ci

Explanation:

The activity of the uranium is determined by using the formula,  

R = 0.693 N/t1/2 (i)

The number of atoms is, N = nNA

Here, NA is the Avogadro number and n is the number of moles. The value of n is m/M, that is, mass/molecular mass. Now the value of N becomes,  

N = (m/M) NA

The m or mass of uranium given is 50.1 grams, and the molecular mass is 238 g/mol, now putting the values we get,  

N = (50 g/238 g) (6.023 × 10²³) = 1.26 × 10²³

The half-life of 238U from year to second is,  

t1/2 = (4.468 × 10⁸ year) (3.16 × 10⁷ s/ 1 year) = 1.412 × 10¹⁶ s

Substituting the values of t1/2 as 1.412 × 10¹⁶, and 1.26 × 10²³ for N in equation (i) we get,  

R = 0.639 (1.26 × 10²³) / 1.412 × 10¹⁶ s  

= 6.18 × 10⁶ Bq (2.7027 × 10⁻¹¹ Ci/1 Bq)

= 1.68 × 10⁻⁵ Ci

Hence, the activity of the plate is 1.68 × 10⁻⁵ Ci

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

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explanation: