Maharashtra State BoardHSC Science (General) 11th
Advertisement Remove all ads

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

Advertisement Remove all ads
Advertisement Remove all ads
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).

Advertisement Remove all ads

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
  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

Balbharati Chemistry 11th Standard Maharashtra State Board
Chapter 13 Nuclear Chemistry and Radioactivity
Exercises | Q 4. (F) | Page 203
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×