Question

When one atom of the radioactive isotope ²³⁹Pu decomposes in a nuclear reactor, it produces a photon of gamma radiation with a wavelength of 1.590 x 10^-13m.

a) How many kilojoules of energy would be released if one gram of ²³⁹Pu decomposed?

b) Use the de Broglie equation to calculate the wavelength of one atom of ²³⁹Pu travelling at 70.0% the speed of light.

Answer 1

Here,

wavelength of photon = 1.59 *10^-13 m

a)

for one gram of 239 Pu

Number of moles of Pu , n = 1/239

number of Pu atoms , N = (1/239) * 6.022 *10^23

Now ,

total energy produced = number of atoms * energy of 1 photon

total energy produced = (1/239) * 6.022 *10^23 * (6.626 *10^-34 * 3 *10^8 /(1.59 *10^-13))

total energy produced = 3.15 *10^9 J

total energy produced = 3.15 *10^6 kJ

the total energy produced is 3.15 *10^6 kJ

b)

speed , v = 0.7c

for the wavelength is

wavelength = h/(m * v)

wavelength = h *sqrt(1 - (v/c)^2)/(m * v)

wavelength = 6.626 *10^-34 * sqrt(1 - 0.7^2)/(0.7 * 3 *10^8 * 239 * 1.66 *10^-27)

wavelength = 5.68 *10^-18 m

the de broglie wavelength is 5.68 *10^-18 m