Question

You place a 1 g cube of 223Ra into an insulated cup containing 0.100 kg of 18 °C water. 223Ra is an alpha emitter with a half-life of 11.43 days. The mass of a 223Ra is 223.018499u. The mass of 219Rn is 219.009477. The mass of an alpha particle is 4.002602.

(a) How much energy must the water absorb to just begin to boil? The specific heat of water is 4190 J/kg K. You can assume you are at sea level so the boiling point of water is 100 °C.

(b) How much energy is released in a single 223Ra alpha decay?

(c) If no energy is lost to the environment, how many decays must occur before the water just starts boiling?

(d) If no energy is lost to the environment, how much time elapses before the water just starts boiling?

Answer 1

a) Energy obsorbed by water,

Q = m*c*dT

= 0.1*4190*(100-18)

= 34358 J

b) Energy relased in one decay, E = delta_m*c^2

= (223.018499 - (219.009477 + 4.002602))*1.67*10^-27*(3*10^8)^2

= 9.65*10^-13 J

c) no of decays occur, N = Q/E

= 34358/(9.65*10^-13)

= 3.56*10^16

d) decay constant, lamda = 0.693/T/12

= 0.693/11.43

= 0.06063 day^-1

initial no of atoms of Ra,

No = no of moles*Na

= (1/223)*6.023*10^23

= 2.7*10^21

let t is the time taken.

No decays, delta_N = No - N

delta_N = No - No*e^(-lamda*t)

3.56*10^16 = 2.7*10^21*(1 - e^(-0.06063*t))

t = 0.00021747 days

= 0.00021747*24*60*60 s

= 18.8 s