﻿ The nucleus Ne1023 decays by β- emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted from the following data:m(Ne1023) = 22.994466 um(Na1123) = 22.989770 u from Physics Nuclei Class 12 Mizoram Board

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The nucleus ${}_{10}{}^{23}\mathrm{Ne}$ decays by β- emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted from the following data:

The $\mathrm{\beta }-\mathrm{decay}$ of ${}_{10}{}^{23}\mathrm{Ne}$ may be represented as:

Ignoring the rest mass of antineutrino $\left(\overline{\mathrm{v}}\right)$ and electron , we get

Mass defect,

$\therefore$

This energy of 4.3792 MeV, is shared by  pair because, ${}_{11}{}^{23}\mathrm{Na}$ is very massive.

The maximum K.E. of  when energy carried by $\overline{\mathrm{v}}$ is zero.

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The half-life of ${}_{38}{}^{90}\mathrm{Sr}$ is 28 years. What is the disintegration rate of 15 mg of this isotope?

Given,
Half- life of ${}_{38}{}^{90}\mathrm{Sr}$ = 28 years.

Using the formula,

$⇒$

90 g of Sr contains 6.023 x 1023 atoms.

$\therefore$ 15 mg of Sr contains,

Disintegration rate,

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The radionuclide ${}^{11}\mathrm{C}$ decays according to
The maximum energy of the emitted positron is 0.960 MeV. Given the mass values:

Calculate Q and compare it with the maximum energy of the positron emitted.

For the given reaction, mass defect is,

Now, Q-value is ,

which, is the maximum energy of the positron.
We have,

The daughter nucleus is too heavy compared to e+ and v. So, it carries negligible energy (Ed ≈ 0). If the kinetic energy (Ev) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q.
Hence, maximum E
e≈ Q.

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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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Obtain approximately the ratio of the nuclear radii of the gold isotope ${}_{79}{}^{197}\mathrm{Au}$ and the silver isotope ${}_{47}{}^{107}\mathrm{Ag}.$

Using the relation between the radius of nucleus and atomic mass,

Atomic mass of gold, A1 = 197
Atomic mass of silver, A2 = 107

$\therefore$

Now, taking log on both sides,

$⇒$

$⇒$

$⇒$
= 1.23, which is the required ratio of the nucleii.

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The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive ${}_{6}{}^{14}\mathrm{C}$${}_{6}{}^{14}\mathrm{C}$ present with the stable carbon istope ${}_{6}{}^{12}\mathrm{C}.$ When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of ${}_{6}{}^{14}\mathrm{C},$ and the measured activity, the age of the specimen can be approximately estimated. This is the principle of  ${}_{6}{}^{14}\mathrm{C}$ dating used in archaeology.
Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus Valley Civilisation.

Given,
Normal activity, R
0 = 15 decays/min,
Present activity R = 9 decays/min,
Half life, T = 5730 years,
Age t =?

As, activity is proportional to the number of radioactive atoms, therefore,

But,

$\therefore$

$⇒$

$⇒$

But,

Therefore,

is the approximate age.
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