In the nuclear reaction,_1H^2 + _1H^2 to _0n^ | vedclass

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

Question

(B) The $Q$-value of a nuclear reaction is given by the difference between the total binding energy of the products and the total binding energy of th

Vedclass · Accepted Answer

$7.37$

Vedclass · Answer

(B) The $Q$-value of a nuclear reaction is given by the difference between the total binding energy of the products and the total binding energy of the reactants.$Q = (BE_{	ext{products}}) - (BE_{	ext{reactants}})$Here,the reactants are two deuterons $(1H^2)$ and the products are one neutron $(0n^1)$ and one helium-$3$ nucleus $(2He^3)$.The binding energy of a neutron is $0 \ MeV$.So,$Q = (BE(2He^3) + BE(0n^1)) - (2 	imes BE(1H^2))$.Given $Q = 3.27 \ MeV$,$BE(1H^2) = 2.23 \ MeV$,and $BE(0n^1) = 0 \ MeV$.$3.27 = BE(2He^3) + 0 - (2 	imes 2.23)$.$3.27 = BE(2He^3) - 4.46$.$BE(2He^3) = 3.27 + 4.46 = 7.73 \ MeV$.

In the nuclear reaction,$_1H^2 + _1H^2 \to _0n^1 + _2He^3$. If the binding energy of deuteron is $2.23 \ MeV$ and the $Q$-value of the reaction is $3.27 \ MeV$,then the binding energy of $_2He^3$ is ......... $MeV$.

Similar Questions

The graph of binding energy per nucleon $B_N$ versus mass number $A$ is given. In which process will energy be released?

The nucleus having the highest binding energy per nucleon is

Draw the nature of the graph of average binding energy per nucleon against atomic mass number and explain its notable points.

$\gamma$-rays radiation can be used to create an electron-positron pair. In this process of pair production,$\gamma$-rays energy cannot be less than ....... $MeV$.

The energy equivalent of $1 \,g$ of substance is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module