If the total binding energies of ${ }_1^2 H ,{ }_2^4 He ,{ }_{26}^{56} Fe$ and ${ }_{92}^{235} U$ nuclei are $2.22, 28.3, 492$ and $1786 \text{ MeV}$ respectively,identify the most stable nucleus of the following.
- A${ }_{26}^{56} Fe$
- B${ }_1^2 H$
- C${ }_{92}^{235} U$
- D${ }_2^4 He$
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Find the binding energy per nucleon for $^{120}_{50}Sn$. Given: mass of proton $m_{p} = 1.00783 \, U$,mass of neutron $m_{n} = 1.00867 \, U$,and mass of tin nucleus $m_{Sn} = 119.902199 \, U$. (Take $1 \, U = 931 \, MeV$) (in $, MeV$)
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$(c)$ The Principle of Conservation of Energy
Medium
View SolutionThe binding energy per nucleon $v/s$ mass number curve for nuclei is shown in the figure. $w, x, y$ and $z$ are four nuclei indicated on the curve. The process that would release energy is:
Difficult
View SolutionAn alpha particle $\left({ }^{4} He\right)$ has a mass of $4.00300 \ amu$. $A$ proton has a mass of $1.00783 \ amu$ and a neutron has a mass of $1.00867 \ amu$. The binding energy of an alpha particle estimated from these data is closest to: (in $MeV$)
MediumWBJEE 2013
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