Question

Enter your answer in scientific notation. Be sure to answer all parts. Calculate the nuclear binding energy (in J) and the nuclear binding energy per nucleon of 107 Ag 47 (106.905093 amu).

Answer 1

The nuclear binding energy =
`1.466* 10^(-13) KJ`

The binding energy per nucleon =1.37×10⁻¹⁵ KJ/nucleon

Step-by-step explanation:

Given:

Number of protons = 47

Number of neutrons = 107-47 = 60

Now,

the mass defect (m)= Theoretical mass - actual mass

m =
`47 (1.007825) + 60(1.008665) - 106.9051`

since,

mass of proton = 1.007825 amu

Mass of neutron = 1.008665 amu

thus,

m =
`47.367775 + 60.5199 - 106.9051`

or

m =
`0.982575 amu`

also

1 amu =
`1.66* 10^(-27) kg`

therefore,

m =
`0.982575 amu * 1.66* 10^(-27) kg`

or

m =
`1.6310745* 10^(-27) kg`

now,

Energy = mass × (speed of light)²

thus,

Energy =
`1.6310745* 10^(27) kg ( 3* 10^8 ms^(-1) )`

or

Energy =
`14.66* 10^(-11) kgm^2/s^2`

or

Energy =
`1.466* 10^(-10) J` =
`1.466* 10^(-13) KJ`

Therefore the nuclear binding energy =
`1.466* 10^(-13) KJ`

Now,

the binding energy per nucleon =
`(1.466* 10^(-13) KJ)/(107)` = 1.37×10⁻¹⁵ KJ/nucleon