Chemistry for Engineers

Question

Directions: Complete the nuclear reaction equation.

1. 21884Po  4 2He + _____________

2. 25399Es + 4 2He  256101Md + _____________ 3. 14261Pm + ____________  14260Nd 4. 21284Po  4 2He + _______________

Directions: Answer the following problem.

1. 206 82Pb  202 80Cm + _____________ 2. 234 90Th  _____________ + 0 -1 3. 4020Ca  40 19K + ___________

4. Write an equation to represent the decay of thorium-230 by alpha emission.

Answer 1

In nuclear equations the law of conservation of mass applies. So in any equation both sides must be equal. If you have alpha decay or emission an alpha particle or a ⁴₂He is produced. The 4 being mass# (protons + neutrons) and the 2 is the atomic # or the number of protons.

So for ²¹⁸₈₄Po--> ⁴₂ He + ²¹⁴₈₂ Pb.

look at the top numbers 218 = 4 + 214

Look at the e bottom numbers 84 = 2 + 82.

It is basic subtraction. The totals on both sides need to be equal. Once you have the atomic number for the product use the periodic table to find the element.

²⁵³₉⁰₉Es + ⁴₂ He --> ²⁵⁶₁₀₁Md + _____________

253 + 4 = 256 + ? 1

99 + 2 = 101 + ? 0

In this case the particle would be ¹₀ n which is simply a neutron.

¹⁴²₆₁Pm + ____________ --> ¹⁴²₆₀Nd

142 + ? = 142. ? = 0

61 + ? = 60 ? = -1. So ⁰_-1 e which is an electron. In this case because it is before the arrow it is electron capture.

There are different nuclear particles

#4 . ²¹²₈₄Po --> ⁴₂ He + ²⁰⁸₈₂ Pb

#1 ²⁰⁶ ₈₂Pb --> ²⁰² ₈₀Cm + ⁴₂ He

#2 ²³⁴ ₉₀Th --> ²³⁴₉₁ Pa + ⁰_-1 e (beta particle)

#3 40 20 Ca --> 40 19K + ⁰₊₁ß (positron)

#4. Write an equation to represent the decay of thorium-230 by alpha emission.

²³⁰₉₀Th --> ⁴₂ He + ²²⁶_{88 Ra}