The binding energy of deuteron {}_1^2H is 1.1 | vedclass

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

Question

(C) The binding energy per nucleon for deuteron $({}_1^2H)$ is $1.112 \, MeV$. Since it has $2$ nucleons,the total binding energy of one deuteron is $

Vedclass · Accepted Answer

$23.8$

Vedclass · Answer

(C) The binding energy per nucleon for deuteron $({}_1^2H)$ is $1.112 \, MeV$. Since it has $2$ nucleons,the total binding energy of one deuteron is $2 	imes 1.112 \, MeV = 2.224 \, MeV$.In the reaction ${}_1^2H + {}_1^2H 	o {}_2^4He + Q$,the initial total binding energy is $2 	imes (2.224 \, MeV) = 4.448 \, MeV$.The $\alpha$-particle $({}_2^4He)$ has $4$ nucleons,and its binding energy per nucleon is $7.047 \, MeV$. Thus,the total binding energy of the final product is $4 	imes 7.047 \, MeV = 28.188 \, MeV$.The energy released $Q$ is the difference between the total binding energy of the products and the reactants: $Q = [B.E.]_{final} - [B.E.]_{initial}$.$Q = 28.188 \, MeV - 4.448 \, MeV = 23.74 \, MeV$.Rounding to the nearest provided option,$Q \approx 23.8 \, MeV$.

The binding energy of deuteron ${}_1^2H$ is $1.112 \, MeV$ per nucleon and an $\alpha$-particle ${}_2^4He$ has a binding energy of $7.047 \, MeV$ per nucleon. Then in the fusion reaction ${}_1^2H + {}_1^2H \to {}_2^4He + Q$,the energy $Q$ released is ......... $MeV$.

Similar Questions

Two lenses in contact made of materials with dispersive powers in the ratio $2 : 1$,behave as an achromatic convex lens of focal length $10\, cm$. The individual focal lengths of the lenses are

If for some $\alpha \in R$,the lines $L_{1}: \frac{x+1}{2}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $L_{2}: \frac{x+2}{\alpha}=\frac{y+1}{5-\alpha}=\frac{z+1}{1}$ are coplanar,then the line $L_{2}$ passes through the point

Which of the following is the composition of glass?

Calculate the $emf$ of the cell $Cu_{(s)}|Cu^{2+}_{(aq)}||Ag^{+}_{(aq)}|Ag_{(s)}$. Given: $E_{Cu^{2+} / Cu}^0 = +0.34 \ V$,$E_{Ag^{+} / Ag}^0 = 0.80 \ V$.

Two point charges $-q$ and $+q$ are located at points $(0, 0, -a)$ and $(0, 0, a)$ respectively. The potential at a point $(0, 0, z)$ where $z > a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module