The isotopic mass of {}_{3}^{7}Li is 7.016005 | vedclass

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

Question

(B) The binding energy $(BE)$ is calculated using the mass defect $(\Delta m)$: $\Delta m = [Z 	imes M_{H}   N 	imes M_{n}] - M_{nucleus}$ Here,$Z =

Vedclass · Accepted Answer

$39.2$

Vedclass · Answer

(B) The binding energy $(BE)$ is calculated using the mass defect $(\Delta m)$: $\Delta m = [Z 	imes M_{H}   N 	imes M_{n}] - M_{nucleus}$ Here,$Z = 3$ (protons),$N = 4$ (neutrons),$M_{H} = 1.007825 \ u$,$M_{n} = 1.008665 \ u$,and $M_{Li} = 7.016005 \ u$. $\Delta m = [3 	imes 1.007825   4 	imes 1.008665] - 7.016005$ $\Delta m = [3.023475   4.034660] - 7.016005 = 7.058135 - 7.016005 = 0.04213 \ u$ $BE = \Delta m 	imes 931 \ MeV/u = 0.04213 	imes 931 \approx 39.2 \ MeV$.

List-$I$ (Nuclear reactor component)	List-$II$ (Used substance)
$1$. Moderator	$(a)$ Uranium
$2$. Control rods	$(b)$ Graphite
$3$. Fuel rods	$(c)$ Boron
$4$. Coolant	$(d)$ Lead
	$(e)$ Sodium

The isotopic mass of ${}_{3}^{7}Li$ is $7.016005 \ u$ and those of the $H$ atom and neutron are respectively $1.007825 \ u$ and $1.008665 \ u$. Then,the binding energy of the ${}_{3}^{7}Li$ nucleus is: (in $MeV$)

Similar Questions

Match List-$I$ and List-$II$ and choose the correct option using the codes given below:

List-$I$ (Nuclear reactor component) List-$II$ (Used substance)

$1$. Moderator $(a)$ Uranium

$2$. Control rods $(b)$ Graphite

$3$. Fuel rods $(c)$ Boron

$4$. Coolant $(d)$ Lead

$(e)$ Sodium

In a nuclear fission chain reaction,a uranium atom undergoes fission to form two different products. The total mass of these products is:

The reaction $_1H^2 + _1H^3 \to _2He^4 + _0n^1 + \text{energy}$ represents

$A$ nuclear reaction must be balanced in terms of

$A$ nuclear reaction is accompanied by a loss of mass equivalent to $0.01864 \; amu$. The energy liberated is .......... $MeV$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module