﻿ Two stable isotopes of lithium Li36 and Li37 have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium. from Physics Nuclei Class 12 Mizoram Board

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`Class 10` `Class 12` Boron has two stable isotopes,  Their respective masses are 10.01294 u and 11.00931 u, and the atomic mass of boron is 10.81u. Find the abundances of

Let, the relative abundance of ${}_{5}{}^{10}\mathrm{B}$ be x%

$\therefore$  Relative abundance of

Atomic weight = weighted average of the isotopes,

$⇒$

i.e.,

866 Views

The three stable isotopes of neon:  have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon.

Given, the masses of isotopes of neon are, 19.99u, 20.99u and 21.99u.
Relative abundance of the isotopes are 90.51 %, 0.27 % and 9.22 % .
Therefore,
Average atomic mass of neon is,

=$\frac{2017.7}{100}$ = 20.17 u

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A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of  (of mass 62.92960 u).

Given, mass of the coin = 3.0 g

Mass of atom = 62.92960 u

Each atom of the copper contains 29 protons and 34 neutrons.

Mass of 29 electrons = 29 x 0.000548 u
= 0.015892 u

Mass of nucleus  = (62.92960 - 0.015892) u

= 62.913708 u

Mass of 29 protons = 29 x 1.007825 u
= 29.226925 u

Mass of 34 neutrons = 34 x 1.008665 u

= 34.29461 u
Total mass of protons and neutrons  = (29.226925 + 34.29461) u
= 63.521535 u

Binding energy  = (63.521535 - 62.913708) x 931.5 MeV
= 0.607827 x 931.5 MeV

$\therefore$ Required energy =

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# Two stable isotopes of lithium  have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium.

Given,
Mass of isotope 1 = 6.01512 u
Mass of isotope 2 = 7.01600 u
Abundance of isoptope 1 = 7.5 %
Abundance of isotope 2 = 92.5 %

Atomic mass of Li =  weighted average of the isotopes

= $\frac{6.01512×7.5+7.01600×92.5}{\left(7.5+92.5\right)}$

=

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Obtain the binding energy (in MeV) of a nitrogen nucleus

${}_{7}{}^{14}\mathrm{N}$ nucleus is made up of 7 protons and 7 neutrons.

Mass of nucleons forming nucleus = 7mp + 7m

= Mass of 7 protons + Mass of 7 neutrons

= 7 × 1.00783 + 7 × 1.00867 u

= 7.05431 + 7.06069 u

= 14.11550.u

Mass of nucleus, mN = 14.00307 u

Therefore,

Mass defect = 14.11550 - 14.00307 = 0.11243 a.m.u.

Energy equivalent to mass defect = 0.11243 × 931 = 104.67 MeV

∴   Binding energy = 104.67 MeV.
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