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Chapter 36: Q. 70 (page 1062)

The radioactive element radium (Ra) decays by a process known as alpha decay, in which the nucleus emits a helium nucleus. (These high-speed helium nuclei were named alpha particles when radioactivity was first discovered, long before the identity of the particles was established.) The reaction is Ra226鈫�Rn222+He4, where Rn is the element radon. The accurately measured atomic masses of the three atoms are 226.0254u, 222.0176u, and 4.0026u. How much energy is released in each decay? (The energy released in radioactive decay is what makes nuclear waste 鈥渉ot.鈥�)

Short Answer

Expert verified

The energy will released is4Mev.

Step by step solution

01

Given information, 

We have given,

the reaction,

Ra226鈫�Rn222+He4

Mass of Rn =222.0176u

Mass of Ra =226.0254u

Mass of He =4.0026u

We have to find the energy is released in the reaction.

02

Simplify

The mass difference will be,

鈭�m=Mass of reactant- Mass of product

鈭�m=226.0254u-(222.0176u+4.0026u)鈭�m=0.0043u

Then Energy will be,

E=鈭�mc2E=0.0043脳931.5E=4Mev

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