0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

Find the energy liberated in the reaction
223Ra209Pb+14C
the atomic masses needed are as follows
223Ra209Pb14C
22.018u208.981u14.003u

Solution
Verified by Toppr

223Ra=223.018u,209Pb=208.981u;14C=14.003u
223Ra209Pb+14C
m=mass223Ramass(209Pb+14C)
=223.018(208.981+14.003)=0.034
energy=M×u=0.034×931=31.65MeV

Was this answer helpful?
4
Similar Questions
Q1
Find the energy liberated in the reaction
223Ra209Pb+14C
the atomic masses needed are as follows
223Ra209Pb14C
22.018u208.981u14.003u
View Solution
Q2

Find the energy liberated in the reaction: 223Ra209Pb+14C
The atomic masses needed are as follows:
223Ra223.018u 209Pb208.981u 14C14.003u


View Solution
Q3

Find the energy liberated in the reaction: 223Ra209Pb+14C
The atomic masses needed are as follows:
223Ra223.018u 209Pb208.981u 14C14.003u


View Solution
Q4
A nuclear reaction along with the masses of the particles taking part in it is as follows:
A+BC+D+QMeV
1.002amu+1.004amu1.001amu+1.003amu+Q
The energy Q liberated in the reaction is
View Solution
Q5
Nuclear binding energy is the energy released during the hypothetical formation of the nucleus by the condensation of individual nucleons. Thus, binding energy per nucleon =Total binding energyNumber of nucleons For example, the mass of hydrogen atom is equal to the sum of the masses of a proton and an electron. For other atoms, the atomic mass is less than the sum of the masses of protons, neutrons and electrons present. This difference in mass, termed as mass defect, is a measure of the binding energy of protons and neutrons in the nucleus. The mass-energy relationship postulated by Einstein is expressed as: ΔE=Δmc2
Where ΔE is the energy liberated, Δm is the loss of mass, and c is the speed of light.
In the reaction 21H+31H42He+10n, if binding energies 21H,31H and 42He are respectively a, b and c (in MeV), then the energy released in this reaction is:
View Solution