The mass defect in a particular nuclear react | vedclass

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

Question

(D) Given mass defect,$\Delta m = 0.3 \,g = 0.3 	imes 10^{-3} \,kg = 3 	imes 10^{-4} \,kg$. Using Einstein's mass-energy equivalence principle,$E =

Vedclass · Accepted Answer

$7.5 	imes 10^6$

Vedclass · Answer

(D) Given mass defect,$\Delta m = 0.3 \,g = 0.3 	imes 10^{-3} \,kg = 3 	imes 10^{-4} \,kg$. Using Einstein's mass-energy equivalence principle,$E = \Delta m c^2$. Substituting the values,$E = (3 	imes 10^{-4} \,kg) 	imes (3 	imes 10^8 \,m/s)^2$. $E = 3 	imes 10^{-4} 	imes 9 	imes 10^{16} = 27 	imes 10^{12} \,J$. To convert Joules to kilowatt-hour $(kWh)$,we divide by $3.6 	imes 10^6 \,J/kWh$. $E = \frac{27 	imes 10^{12}}{3.6 	imes 10^6} \,kWh$. $E = 7.5 	imes 10^6 \,kWh$.

The mass defect in a particular nuclear reaction is $0.3 \,g$. The amount of energy liberated in kilowatt-hour $(kWh)$ is: (Velocity of light $c = 3 \times 10^8 \,m/s$)

Similar Questions

Nuclear binding energy is equivalent to

$A$ nucleus $^{A}_{Z} X$ has mass represented by $M(A, Z)$. If $M_p$ and $M_n$ denote the mass of a proton and a neutron respectively, and $B.E.$ is the binding energy in $MeV$, then:

If an electron and a positron annihilate,the energy released is ........

An alpha particle $\left({ }^{4} He\right)$ has a mass of $4.00300 \ amu$. $A$ proton has a mass of $1.00783 \ amu$ and a neutron has a mass of $1.00867 \ amu$. The binding energy of an alpha particle estimated from these data is closest to: (in $MeV$)

The average binding energy per nucleon in the nucleus of an atom is approximately

Vedclass Test Series

Exam Paper Generator

Online Exam Module