Home
Class 12
PHYSICS
Calculate the average binding energy per...

Calculate the average binding energy per nucleon of `._(41)^(93)Nb` having mass 9.2.906 u..

Text Solution

Verified by Experts

In order to compute binding energy, let us first find the total mass of all protons and neutrosn in Nb and substarct mass of the Nb:
Given: `m_(p) =1.007276 u` and `m_(n)=1.008665 u`
Number of protons: `N_(p) =41`
Number of neutrons: `N_(n) =93 -41=52`
mass differences: `Delta m=41 m_(p) + 52 m_(n) -m_(Nb)`
`=41(1.007825 u) +52(1.008665 u) -(92.9063768 u)`
`=0.865028 u `
Thus, binding energy per nuclear is .
`(E_(b))/(A)=((Delta)c^(2))/(A)=((0.865028u)(931.5MeV//u))/(93)`
`=8. 66 MeV ` nucleon^(-1)`.
Promotional Banner

Topper's Solved these Questions

  • NUCLEAR PHYSICS

    CENGAGE PHYSICS|Exercise Exercise 5.2|27 Videos

  • NUCLEAR PHYSICS

    CENGAGE PHYSICS|Exercise Subjective|35 Videos

  • NUCLEAR PHYSICS

    CENGAGE PHYSICS|Exercise Solved Examples|17 Videos

  • Moving charges and magnetism

    CENGAGE PHYSICS|Exercise Question Bank|20 Videos

  • NUCLEI

    CENGAGE PHYSICS|Exercise QUESTION BANK|59 Videos

Similar Questions

Explore conceptually related problems

The value of binding energy per nucleon is

For atomic nuclei, the binding energy per nucleon

What is the binding energy per nucleon for ""^(120)Sn ?

The binding energy per nucleon is largest for

As the mass number increase binding energy per nucleon :

Calculate the binding energy per nucleon of ._(20)^(40)Ca . Given that mass of ._(20)^(40)Ca nucleus = 39.962589 u , mass of proton = 1.007825 u . Mass of Neutron = 1.008665 u and 1 u is equivalent to 931 MeV .

Average binding energy per nucleon over a wide range is

Calculate the binding energy per nucleon of ._(20)Ca^(40) nucleus. Mass of (._(20)Ca^(40)) = 39.962591 am u .

Find the binding energy per nucleon of _79^197Au if its atomic mass is 196.96 u.