# Calculate the binding energy per nucleon of X3684X236284Kr whose atomic mass is 83.913 u. (Mass of neutron is 1.0087 u and that of H atom is 1.0078 u). - Chemistry

Numerical

Calculate the binding energy per nucleon of $\ce{^84_36Kr}$ whose atomic mass is 83.913 u. (Mass of neutron is 1.0087 u and that of H atom is 1.0078 u).

#### Solution

Given: A = 84, Z = 36, m = 83.913 u, mn = 1.0087 u, mH = 1.0078 u

To find: Binding energy per nucleon (bar"B")

Formulae:

1. ∆m = ZmH + (A - Z)mn - m
2. B.E. = ∆m × 931.4 MeV
3. bar"B" = "B.E."/"A"

Calculation:

1. ∆m = ZmH + (A - Z)mn - m
= (36 × 1.0078) + (48 × 1.0087) - 83.913
= 36.2808 + 48.4176 - 83.913
= 0.7854 u
2. B.E. = ∆m × 931.4 MeV
= 0.7854 × 931.4
= 731.4 MeV (by using log table)
3. bar"B" = "B.E."/"A" = 731.4/84
= 8.706 MeV (by using log table)

Binding energy per nucleon of $\ce{^84_36Kr}$ = 8.706 MeV

Concept: Introduction: Nuclear Chemistry is a Branch of Physical Chemistry
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#### APPEARS IN

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