# 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

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

#### 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