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Calculate the energy in MeV released in the nuclear reaction X77174X2772174Ir⟶X75170X2752170Re+X24X2224He Atomic mass: Ir = 173.97 u, Re = 169.96 u and He 4.0026 u - Chemistry

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Numerical

Calculate the energy in MeV released in the nuclear reaction \[\ce{^174_77Ir -> ^170_75Re + ^4_2He}\]

Atomic mass: Ir = 173.97 u, Re = 169.96 u and He 4.0026 u

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Solution

Given: mIr = 173.97 u, mRe = 169.96 u, mHe = 4.0026 u

To find: Energy released

Formulae:

  1. ∆m = (mass of 174Ir) - (mass of 170Re + mass of 4He)
  2. E = ∆m × 931.4 MeV

Calculation: 

  1. ∆m = (mass of 174Ir) - (mass of 170Re + mass of 4He) 
    = 173.97 - (169.96 + 4.0026)
    = 7.4 × 10-3 u
  2. E = ∆m × 931.4
    = 7.4 × 10-3 × 931.4
    = 6.89236 MeV ≈ 6.892 MeV

The energy released in given nuclear reaction is 6.892 MeV.

Concept: Nuclear Reactions
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APPEARS IN

Balbharati Chemistry 11th Standard Maharashtra State Board
Chapter 13 Nuclear Chemistry and Radioactivity
Exercises | Q 4. (G) | Page 203
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