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Physics Part II

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Class 10 Class 12

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


The β-decay of Ne1023 may be represented as:

            Ne1023  Na1123 - e-10 + v¯ + Q 

Ignoring the rest mass of antineutrino v¯ and electron , we get 

Mass defect,                  
                   m = m(Ne1023) - m (Na1123)         = 22.994466 - 22.989770          = 0.004696 u          

  Q = 0.004696×931 MeV = 4.372 MeV. 

This energy of 4.3792 MeV, is shared by e- and v¯ pair because, Na1123 is very massive.

The maximum K.E. of e- = 4.372 MeV, when energy carried by v¯ is zero.

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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 Cu2963 atoms (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 

 Required energy = 6.023 × 102363×3×0.607827× 931.5 MeV
                              = 1.6 × 1025 MeV = 2.6 ×1012J.

1508 Views

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 C614C614 present with the stable carbon istope C612. 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 C614, and the measured activity, the age of the specimen can be approximately estimated. This is the principle of  C614 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,

                   NN0 = RR0 = 915 

But,             NN0 = e-λt

                e-λt = 915 =35 

                   e+λt = 53

     λt logee = loge53 = 2.3026 log 1.6667

            λt = 2.3026 × 0.2218    = 0.5109  t = 0.5109λ 

But, λ = 0.693T =0.6935730 Yr-1

Therefore,
               t =0.51090.693/5730 = 0.5109 × 57300.693

               t = 4224.3 years. is the approximate age.
134 Views

Obtain approximately the ratio of the nuclear radii of the gold isotope Au79197 and the silver isotope Ag47107.

Using the relation between the radius of nucleus and atomic mass,
                             R  A1/3 

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

                       R1R2 = A1A21/3 

                                = 1971071/3 = (1.84)1/3 

Now, taking log on both sides, 

         log10R1R2 = log10(1.84)1/3 

         log10R1R2 = 13log10(1.84)             

                              = 13×0.2648= 0.08827 

                   R1R2 = antilog(0.08827) 
                              = 1.23, which is the required ratio of the nucleii. 

220 Views

The half-life of Sr3890 is 28 years. What is the disintegration rate of 15 mg of this isotope?

Given,
Half- life of Sr3890 = 28 years. 

Using the formula,
                           λ =0.693T

                       λ = 0.69328×365×24×60×60
                                  
                              = 7.85 × 10-10 s-1 
90 g of Sr contains 6.023 x 1023 atoms.

 15 mg of Sr contains,

N0 = 6.023 × 1023 × 15 × 10-390atoms 

N0 = 1.0038 × 1020 atoms

Disintegration rate, dNdt = -λ N0

                                       = -7.85 × 10-10 × 1.0038 × 1020= 7.88 ×1010 dps or Bq= 7.88 × 10103.7 × 1010Ci

                                    = 2.13 Ci.

292 Views

The radionuclide C11 decays according to C611  B511+e++v;  T1/2 = 20.3 min.
The maximum energy of the emitted positron is 0.960 MeV. Given the mass values:
mC611 = 11.011434 u  and  m B511 = 11.009305 u.
Calculate Q and compare it with the maximum energy of the positron emitted.

For the given reaction, mass defect is, 
                               m =  [m C611 - 6me - m B511 - 5 me+me        =  mC611 - m(B511) -  2 me        =  11.011434 u - 11.009305 u - 2 × 0.000548 u        =  0.001033 u

Now, Q-value is , 

              Q = 0.001033 × 931.5 MeV     = 0.962 MeV 

which, is the maximum energy of the positron. 
We have, 
                     Q = Ed+Ee+Ev 
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.

302 Views