24.143 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Plutonium-242 decays to uranium-238 by emission of an $\alpha$-particle with an energy of 4.853 MeV. The  ²³⁸U that forms unstable and emits a $\gamma$-ray ($\lambda =0.02757 nm$).

(a) Write balanced equations for these reactions.

(b) What would be the energy of the -particle if ²⁴²P decayed directly to the more stable ²³⁸U?

Step-by-Step Solution

Answer

(a) $$\begin{gathered} {}_{94}^{249}Pu{\to }_{92}^{238}U{+}_2^4\alpha \\ {238}_{92}U\to {238}_{92}U+\gamma \end{gathered}$$ 

(b) 4.989 MeV.

 

Step 1: Definition of Half-life

The time it takes for a given reactant's concentration to reach 50% its initial concentration is known as the half-life of a chemical reaction (i.e., the time taken for the reactant concentration to reach half of its initial value).

It's commonly represented in seconds and is denoted by the sign 't_1/2'.

Step 2: Subpart (a)

Balanced equation

$$\begin{gathered} {}_{94}^{249}Pu{\to }_{92}^{238}U{+}_2^4\alpha \\ {238}_{92}U\to {238}_{92}U+\gamma \end{gathered}$$

Step 3: Subpart (b)

To determine the energy in MeV we need to use the formula of $\gamma$ ray.

$E=\frac{hc}{\lambda }$

$$\begin{gathered} E=\frac{hc}{\lambda } \\ =\frac{\left(6.626x{10}^{-34} J.s\times 2.99792x{10}^8 m/s\right)}{\left(0.02747\times {10}^{-9} m \right)} \\ =723.1 J \\ =0.0450 MeV \left(\because 1 MeV=1.60\times {10}^{-13} J\right) \end{gathered}$$

 

Total Energy:

$$\begin{gathered} E_T=4.853MeV+0.0450MeV\backslash \mathrm{hfill} \\ {E}_T=4.989MeV \end{gathered}$$