The figure shows a plot of binding energy per | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Energy is released $(\varepsilon  >  0)$ in a nuclear reaction if the total binding energy of the products is greater than the total binding energ

Vedclass · Accepted Answer

$(i)$ and $(iv)$

Vedclass · Answer

(A) Energy is released $(\varepsilon  >  0)$ in a nuclear reaction if the total binding energy of the products is greater than the total binding energy of the reactants.$1$. For reaction $(i)$, $A + B 	o C + \varepsilon$: The nucleus $C$ has a higher binding energy per nucleon than $A$ and $B$. Thus, the total binding energy of the product $C$ is greater than the sum of the binding energies of $A$ and $B$. Therefore, $\varepsilon  >  0$.$2$. For reaction $(iv)$, $F 	o D + E + \varepsilon$: The nuclei $D$ and $E$ have higher binding energy per nucleon than $F$. Thus, the total binding energy of the products $D$ and $E$ is greater than the binding energy of the reactant $F$. Therefore, $\varepsilon  >  0$.Hence, $\varepsilon$ is positive in reactions $(i)$ and $(iv)$.

Similar Questions

One million electron volt $(1\,MeV)$ is equal to

If the binding energy per nucleon in $_3^7Li$ and $_2^4He$ nuclei are $5.60 \, MeV$ and $7.06 \, MeV$ respectively,then in the reaction $p + {}_3^7Li \to 2 {}_2^4He$,the energy of the proton must be ........... $MeV$.

Let ${m_p}$ be the mass of a proton,${m_n}$ the mass of a neutron,${M_1}$ the mass of a $_{10}^{20}Ne$ nucleus,and ${M_2}$ the mass of a $_{20}^{40}Ca$ nucleus. Then:

Binding energy per nucleon of ${ }_1^2 H$ and ${ }_2^4 He$ are $1.1 \ MeV$ and $7.0 \ MeV$ respectively. Energy released in the process ${ }_1^2 H + { }_1^2 H \rightarrow { }_2^4 He$ is: (in $MeV$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module