Question

Be sure to answer all parts. Plutonium−242 decays to uranium−238 by emission of an α particle with an energy of 4.853 MeV. The 238U that forms is unstable and emits a γ ray (λ = 0.02757 nm). (a) Write a balanced net equation for these reactions. → (b) What would be the energy of the α particle if 242Pu decayed directly to the more stable 238U? MeV

Answer 1

Answer:

(a)²⁴²₉₄Pu ⇒ ⁴₂He + ²³⁸₉₂U ⇒ ²³⁸₉₂U + ⁰₀γ

(b) 4.898 MeV

Explanation:

In the nuclear reaction, it was stated that plutonium-242 decayed firstly to uranium-238 and alpha, and lastly to a stable uranium-238 by emitting a gamma ray. The balanced equation for the nuclear reactions is shown below:

(a)²⁴²₉₄Pu ⇒ ⁴₂He + ²³⁸₉₂U ⇒ ²³⁸₉₂U + ⁰₀γ

(b) The energy emitted by releasing a gamma ray is calculated using:

E = hc/λ

where

h = 4.136*10^-15 eV.s

c = 299792458 m/s

λ = 0.02757 nm = 0.02757*10^-9 m

Therefore:

E = (4.136*10^-15)*(299792458)/0.02757*10^-9 = 44974.31 eV

The total energy if the stable 238U was produced directly would be

4.853*10^6 + 44974.31 = 4.898 MeV