Question

unit d: energy and the environment
harvesting nuclear energy
remember to use all proper notations (\\(_{z}^{a}x\\)) in your answer. this assignment is best done by hand.
alpha radiation

1. write the balanced nuclear equation showing the alpha decay for each isotope given

a. beryllium-8

b. uranium-232

1. each of the following atoms is a product of an alpha-decay reaction. write a balanced nuclear equation for the reaction. what is the reactant?

a. uranium-235

1. radium-226 is an unstable isotope that decays to radon-222

a. write the balanced nuclear equation for this process

b. identify the type of radiation produced by the decay of radium-226

Explanation:

1) a. Beryllium - 8 Alpha Decay

Step1: Recall alpha decay notation

An alpha particle is $_{2}^{4}\text{He}$. For beryllium - 8 ($_{4}^{8}\text{Be}$), in alpha decay, the mass number decreases by 4 and the atomic number decreases by 2.

Step2: Write the equation

The reactant is $_{4}^{8}\text{Be}$, the product will have mass number $8 - 4=4$ and atomic number $4 - 2 = 2$ (which is helium, $_{2}^{4}\text{He}$? Wait, no, wait. Wait, beryllium - 8 alpha decay: $_{4}^{8}\text{Be}
ightarrow_{2}^{4}\text{He}+_{2}^{4}\text{He}$? Wait, no, wait, beryllium - 8: atomic number 4, mass number 8. After alpha decay (losing 2 protons and 2 neutrons, mass number 4, atomic number 2), so the product should be two alpha particles? Wait, no, actually, beryllium - 8 undergoes alpha decay to form two helium - 4 nuclei. So the balanced equation is $_{4}^{8}\text{Be}
ightarrow2_{2}^{4}\text{He}$? Wait, no, let's check mass and charge balance. Left side: mass number 8, atomic number 4. Right side: 2*4 = 8 (mass), 2*2 = 4 (atomic number). So yes, $_{4}^{8}\text{Be}
ightarrow2_{2}^{4}\text{He}$.

1) b. Uranium - 232 Alpha Decay

Step1: Recall uranium - 232 has atomic number 92, mass number 232. Alpha particle is $_{2}^{4}\text{He}$.

Step2: Calculate product's mass and atomic number

Mass number of product: $232-4 = 228$. Atomic number of product: $92 - 2=90$ (which is thorium, Th). So the equation is $_{92}^{232}\text{U}
ightarrow_{90}^{228}\text{Th}+_{2}^{4}\text{He}$.

2) a. Uranium - 235 as Alpha Decay Product

Step1: Recall alpha decay: reactant has mass number = product mass number + 4, atomic number = product atomic number+2.

Step2: For uranium - 235 ($_{92}^{235}\text{U}$), the reactant will have mass number $235 + 4=239$ and atomic number $92+2 = 94$ (which is plutonium, Pu). So the equation is $_{94}^{239}\text{Pu}

ightarrow_{92}^{235}\text{U}+_{2}^{4}\text{He}$. The reactant is plutonium - 239 ($_{94}^{239}\text{Pu}$).

3) a. Radium - 226 Decay to Radon - 222

Answer:

s:

1. a. $\boldsymbol{_{4}^{8}\text{Be}

ightarrow2_{2}^{4}\text{He}}$

1. b. $\boldsymbol{_{92}^{232}\text{U}

ightarrow_{90}^{228}\text{Th}+_{2}^{4}\text{He}}$

1. a. Reactant: $\boldsymbol{_{94}^{239}\text{Pu}}$, Equation: $\boldsymbol{_{94}^{239}\text{Pu}

ightarrow_{92}^{235}\text{U}+_{2}^{4}\text{He}}$

1. a. $\boldsymbol{_{88}^{226}\text{Ra}

ightarrow_{86}^{222}\text{Rn}+_{2}^{4}\text{He}}$

1. b. $\boldsymbol{\text{Alpha radiation}}$