In 2006, an ex-KGB agent was murdered in London. Subsequent investigation showed

Question

In 2006, an ex-KGB agent was murdered in London. Subsequent investigation showed that the cause of death was poisoning with the radioactive isotope 210 Po, which was added to his drinks/food. (a) 210 Po is prepared by bombarding 209 Bi with neutrons. Write an equation for the reaction. (b) Who discovered the ele-ment polonium? ( Hint: See Appendix 1.) (c) The half-life of 210 Po is 138 d. It decays with the emission of an particle. Write an equation for the decay process. (d) Calculate the energy of an emitted particle. Assume both the parent and daughter nuclei to have zero kinetic energy. The atomic masses are: 210 Po (209.98285 amu), 206 Pb (205.97444 amu), 42 (4.00150 amu). (e) Ingestion of 1 µ g of 210 Po could prove fatal. What is the total energy released by this quantity of 210 Po?

Explanation / Answer

Solution:- (a) Neutron is represented as 0n1. From the symbol it is clear that if neutron bombardment takes place then mass number increases by one unit and the atomic number remains the same. Atomic number of Bi is 83. So, if neutron is bombarded on it then atomic number will still be 83 but mass number will change from 209 to 210. Atomic number of Po is 84. So to form Po from Bi we need to do two steps. In first steps mass number increases by one unit and in second step which is beta decay, atomic number increases by one unit. The equations are written as...

83Bi209 + 0n1 ---------> 83Bi210

83Bi210 + -1e0 ----------> 84Po210

(b) It was Marie Curie who discovered it with her husband Pierre Curie and it was named Polonium on the basis of Marie Curie's homeland "Poland".

(c) In alpha decay, atomic number decreases by 2 units and mass number decreases by 4 units. Atomic number of Po is 84 so as the atomic number decreases by 2 units so the new atomic number will be 82 and the element with atomic number 82 is Pb. Mass number of Po is given as 210. Since the mass number decreases by 4 units so the new mass number will be 206. The equation could be written as...

84Po210    ----------> 82Pb206 + 2He4(alpha particle)

(d) from Einstein energy equation, delta E = delta m.c2

where delta m is mass defect which is the difference of sum of masses of products and recatants and c is the speed of light.

delta m = (205.97444 + 4.00150) - 209.98285

delta m = -0.00691 amu

we need to convert amu into kg. 1 amu = 1.6605 x 10-27 kg

So, 0.00691 amu x (1.6605 x 10-27 kg/ 1 amu) = 1.15 x 10-29 kg

E = (1.15 x 10-29 kg) (3.0 x 108 m/s)2 = 1.04 x 10-12 J         

So, the energy of an emitted alpha particle is 1.04 x 10-12 J.

(e) we will use the unit conversions for this and take help of the energy for each alpha particle that is calculated in part d. we will convert 1 microgram into grams and these grams will be converted into the moles of Po. there is 1:1 mol ratio between Po and alpha particles. So, we can calculate alpha particles and then with the help of calculated energy for one alpha particle we will calculate the total energy released.

1 microgram x (10-6 g/1 microgram) x (1mol of Po/209.98285 g) x (1 mol of alpha particle/ 1mol of Po) x (6.022 x 1023 alpha particles/1 mol of alpha particle) x (1.04 x 10-12 J/ 1 alpha particle) = 2.98 x 103 J or 2.98 kJ

So, total energy released is 2.98 x 103 J or 2.98 kJ.

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