Question

Calculate the binding energy per nucleon for $_{10}^{20}Ne$, $_{26}^{56}Fe$ and $_{92}^{238}U$. Given that mass of a neutron is$1.008665amu$, the mass of a proton is$1.007825amu$, the mass of \[_{10}^{20}Ne\] is $19.9924amu$, the mass of $_{26}^{56}Fe$ is $55.93492amu$, $_{92}^{238}U$ is $238.050783amu$.

Answer 1

Hint : In this question, we calculate binding energy per nucleon for$_{10}^{20}Ne$, $_{26}^{56}Fe$ and $_{92}^{238}U$. To calculate the binding energy we have to calculate mass defect first. After calculating mass defect we use this formula for the calculation of binding energy $BE=\dfrac{\Delta m{{c}^{2}}}{A}$.

Complete step by step answer:
Given:
 $Mass\text{ }of\text{ }neutron=1.008665amu$
 $Mass\text{ }of\text{ }proton=1.007825amu$
 $Mass\text{ }of~_{10}^{20}Ne=19.9924amu$
 $Mass\text{ }of_{26}^{56}Fe=55.93492amu$
 $Mass\text{ }of_{92}^{238}U=238.050783amu$
  We calculate mass defect for$_{10}^{20}Ne$. The formula for the calculation of the mass defect is given as,
 $\Delta m=\left[ 10{{m}_{p}}+(A-Z){{m}_{n}} \right]+{{M}_{Ne}}$
Here
 $\Delta m=\text{Mass defect}$
 ${{m}_{p}}=\text{Mass of proton}$
 ${{m}_{n}}=\text{Mass of neutron}$
 ${{M}_{Ne}}=\text{Mass of }_{10}^{20}\text{Ne}$
Now put the value in the above equation,
 $\Rightarrow \Delta m=[10\times 1.007825+10\times 1.008665]-19.9924$
We simplify this equation
 $\Rightarrow \Delta m=[10.07825+10.08665]-19.9924$
On further solving
 $\Rightarrow \Delta m=20.1649-19.9924$
Mass defect we get
 $\Delta m=0.1725amu$
Now we calculate Binding energy per nucleon,
 $BE=\dfrac{\Delta m{{c}^{2}}}{A}$
Here we put the values in the equation,
 $\Rightarrow BE=\dfrac{0.1725{{c}^{2}}}{20}=0.0086{{c}^{2}}amu$
We change $amu$ into \[MeV\]. $1amu=931.5MeV/{{c}^{2}}$
 \[\Rightarrow 0.0086\times 931.5=8.03MeV\]
We calculate mass defect for\[_{26}^{56}Fe\].
 \[\Rightarrow \Delta m=[26{{m}_{p}}+(56-26){{m}_{n}}]+{{M}_{Fe}}\]
Now we put values in the equation
 \[\Rightarrow \Delta m=[26\times 1.007825+30\times 1.008665]-55.93492\]
After simplifying
 \[\Rightarrow \Delta m=[26.20345+30.25995]-55.93492\]
After further solving
 \[\Rightarrow \Delta m=56.4634-55.93492\]
Here we get the mass defect of Ferrus.
 \[\Delta m=0.5285amu\]
Now we calculate Binding energy per nucleon,
 $BE=\dfrac{\Delta m{{c}^{2}}}{A}$
Here we put the values in the equation
 \[\Rightarrow BE=\dfrac{0.5285{{c}^{2}}}{56}=0.0086{{c}^{2}}amu\]
We change $amu$ into \[MeV\]. $1amu=931.5MeV/{{c}^{2}}$
 \[\Rightarrow 0.0094\times 931.5=8.76MeV\]
We calculate mass defect for$_{92}^{238}U$.
 \[\Delta m=[92{{m}_{p}}+(238-92){{m}_{n}}]+{{M}_{U}}\]
Now we put the values in the equation
 \[\Delta m=[92\times 1.007825+146\times 1.008665]-238.050783\]
After simplifying
 \[\Delta m=[92.7199+147.26509]-238.050783\]
After further solving
 \[\Delta m=239.98499-238.050783\]
Here we get the mass defect of Uranium.
 \[\Delta m=1.934amu\]
Now we calculate Binding energy per nucleon,
 $BE=\dfrac{\Delta m{{c}^{2}}}{A}$
Now we put the values in the equation,
 \[\Rightarrow BE=\dfrac{1.934{{c}^{2}}}{238}=0.0086{{c}^{2}}amu\]
 We change $amu$ into \[MeV\]. $1amu=931.5MeV/{{c}^{2}}$
 \[\Rightarrow 0.0081\times 931.5=7.57MeV\]

Note: For calculation of binding energy of Uranium, Ferrum, Neon we have to find a defect in their mass after that we calculate binding energy per nucleon. To understand this type of question we have to study their mass and other properties uranium is a highly reacted element and it provides a high amount of heat when it starts reacting so these types of elements have many properties.