If the binding energy per nucleon of _1H^2 is | vedclass

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

Question

(B) The reaction is $_1H^2 + _1H^2 	o _2He^4$. Total binding energy of reactants = $(2 	imes 2 	imes x) = 4x$. Total binding energy of product = $(

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) The reaction is $_1H^2 + _1H^2 	o _2He^4$. Total binding energy of reactants = $(2 	imes 2 	imes x) = 4x$. Total binding energy of product = $(4 	imes 7x) = 28x$. Energy liberated $(Q)$ = $(	ext{Binding energy of product} - 	ext{Binding energy of reactants}) = 28x - 4x = 24x$. The ratio of energy liberated to the binding energy per nucleon of $_1H^2$ is $\frac{24x}{x} = 24$.

If the binding energy per nucleon of $_1H^2$ is $x$ and that of $_2He^4$ is $7x$,then the ratio of the energy liberated in the reaction $_1H^2 + _1H^2 \to _2He^4$ to the binding energy per nucleon of $_1H^2$ is:

Similar Questions

The number of neutrons in the parent nucleus which gives $N^{14}$ on $\beta$-emission and the parent nucleus is

The phenomenon of radioactivity arises from the

The positron is

The order of density in the nucleus is:

In nuclear reactors,the speed of the neutrons is slowed down by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module