Radioactivity

Chemistry Level 2

Calculate the number of $\alpha$ and $\beta$ particles emitted by the conversion of $^{238}_{92}U\rightarrow^{206}_{82}Pb$

Let the number of $\alpha$ and $\beta$ particles be $x$ and $y$ respectively. Find $x+y$ .

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The answer is 14.

2 solutions

Ronald Keswantono
Aug 30, 2015

238 - 4x = 206 4x = 32 x=8

92-8(2)+y=82 y = 6

x+y = 14

Melissa Quail
Sep 3, 2015

An alpha particle consists of 2 protons and 2 neutrons so for each alpha particle emitted the number of protons and the number of neutrons each decrease by 2. A beta particle consists of an electron but in order to create the electron, a neutron breaks down into a proton and an electron so each beta particle emitted increases the number of protons by 1 and decreases the number of neutrons by 1. 92 - 82 = 10 so the proton number has decreased by 10. 238 - 206 = 32 so the total number of protons and neutrons has decreased by 32, therefore the number of neutrons has decreased by 22. 12 more neutrons have been lost than protons. Alpha particles release equal numbers of each so beta particles have caused this difference. Each beta particle increases the difference by 2 so there were 6 beta particles emitted. 6 beta particles would have created a +6 change in the number of protons but there was a -10 change so 16 protons were lost and therefore there were 8 alpha particles emitted. 6 + 8 = 14 so the answer is 14.

Radioactivity

Image Credit: Wikimedia Stannered .

The answer is 14.

2 solutions

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