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Determine Q for the reactionL37i+H11B47e+n01

Short Answer

Expert verified

Energy releasedQ=-1.64MeV

Step by step solution

01

Given data and concept

The reaction between lithium and hydrogen:

Li37+H11Be47+n01

  • mL37=6.941u- the mass of lithium
  • mH11=1.0079u- the mass of hydrogen
  • mBe47=7.01692u- the mass of beryllium
  • mn02=1.00866u- unknown

So, we had to determine the released heat Q for the reaction.

We will be applying an equation that determines energy as:

Q=mc2

Where:m - mass defect C - speed of light

Where the mass defect mis given by: m=mL37+mH11-mBe47-mn02c2.

02

Put known values for finding Q

For calculating the Q for the reaction:

Q=(6.941u+1.0079u-7.01692u-1.00866u)c2=-0.00176931.5MeV=-1.64MeV

Therefore, the amount of energy released isQ=-1.64MeV

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