The binding energy per nucleon for a deuteron | vedclass

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

Question

(D) The given nuclear reaction is: $_{1}^{2}H + {}_{1}^{2}H 	o {}_{2}^{4}He + Q$.The binding energy per nucleon for $_{1}^{2}H$ is $x_1$. Since there

Vedclass · Accepted Answer

$4(x_2 - x_1)$

Vedclass · Answer

(D) The given nuclear reaction is: $_{1}^{2}H + {}_{1}^{2}H 	o {}_{2}^{4}He + Q$.The binding energy per nucleon for $_{1}^{2}H$ is $x_1$. Since there are $2$ nucleons in a deuteron,the total binding energy of one deuteron is $2x_1$.The binding energy per nucleon for $_{2}^{4}He$ is $x_2$. Since there are $4$ nucleons in an $\alpha -$ particle,the total binding energy of one $\alpha -$ particle is $4x_2$.The energy released $(Q)$ in a nuclear reaction is equal to the difference between the total binding energy of the products and the total binding energy of the reactants.$Q = (	ext{Total B.E. of products}) - (	ext{Total B.E. of reactants})$$Q = (4x_2) - (2x_1 + 2x_1)$$Q = 4x_2 - 4x_1 = 4(x_2 - x_1)$.

The binding energy per nucleon for a deuteron $(_{1}^{2}H)$ and an $\alpha -$ particle $(_{2}^{4}He)$ are $x_1$ and $x_2$ respectively. The energy $(Q)$ released in the reaction $_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{4}He + Q$ is

Similar Questions

The mass defect of $ { }_{2}^{4} He $ is $ 0.03 \ u $. The binding energy per nucleon of helium (in $ MeV $ ) is

State Einstein's special theory of relativity and provide the mass-energy equivalence formula.

The 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:

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

The energy released in a nuclear reaction is due to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module