Nucleus A has a mass number of 220 and its bi | vedclass

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

Question

(D) The energy released in a nuclear fission reaction is given by the difference between the total binding energy of the products and the total bindin

Vedclass · Accepted Answer

$176$

Vedclass · Answer

(D) The energy released in a nuclear fission reaction is given by the difference between the total binding energy of the products and the total binding energy of the reactant.Total binding energy of the reactant (Nucleus $A$): $BE_A = 220 	imes 5.6 \, MeV = 1232 \, MeV$.Total binding energy of the products (Nuclei $B$ and $C$): $BE_{B+C} = (105 	imes 6.4) + (115 	imes 6.4) \, MeV = (105 + 115) 	imes 6.4 \, MeV = 220 	imes 6.4 \, MeV = 1408 \, MeV$.The energy $Q$ released is:$Q = BE_{products} - BE_{reactant}$$Q = 1408 \, MeV - 1232 \, MeV = 176 \, MeV$.

Similar Questions

The mass defect in a particular reaction is $0.4 \,g$.
The amount of energy liberated is $n \times 10^7 \,kWh$,where $n =$ . . . . . . . (speed of light $= 3 \times 10^8 \,m/s$)

Calculate the energy equivalent of $1 \ g$ of substance.

The masses of a proton, neutron, and helium nucleus are $1.0073\,u$, $1.0087\,u$, and $4.0015\,u$ respectively. The binding energy of the helium nucleus is $.........\,MeV$.

The plot of binding energy per nucleon against the mass number for stable nuclei is shown in the figure. Which curve is correct?

$A$ nucleus with mass number $242$ and binding energy per nucleon as $7.6\,MeV$ breaks into two fragments,each with mass number $121$. If each fragment nucleus has a binding energy per nucleon of $8.1\,MeV$,the total gain in binding energy is $........MeV$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module