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A radioactive nucleus decays with the activity shown in the graph above. What is the half life of the nucleus? (in minutes)
use log_{10 }2 = 0.3 & log_{10}e = 0.43
Count at t = 0 is 6 × 10^{3}
Count at t = 30 is 3 × 10^{2}
N_{0} = 6 × 10^{3}
The correct answer is: 7
If the total angular momentum quantum number of a nucleus (A = 200) is J = 1, and if it is due to the rotation of the nucleus as a rigid body, find out the frequency. (in 10^{12} rad/s)
A = 200
mass = 200 × 1.6 × 10^{–27} kg
J = 1
⇒ Angular momentum =
(for a rigid body with moment of inertia = I)
(solid sphere since is A is quite high)
The correct answer is: 2.3
Plutonium ^{238}Pu_{94} has an α decay half life of 90 years (2.7 × 10^{9} s). If each of the α particles is emitted with 5.5 MeV energy, then find the average power released (till 50% has decayed) if there are 238 gms of ^{238}Pu (i.e. 6 × 10^{23} atoms)(in Watts)
Solution=
t_{1/2}=2.7x10^{9}s
avg power=energy/time
=(No. of atoms decayed x 5.5MeV)/t_{1/2}
=N_{0}x5.5MeV/t_{1/2}
=3x10^{23}x5.5MeV/2.7x10^{9}
=(11.6x10^{23}x10^{+16}/10^{9})x10^{19}
≈116W
The correct answer is 116W
If the energy of the α particle emitted by ^{231}Am is 5.48 MeV, find the distance of closest approach between the α particle and ^{197}Au nucleus (in 10^{–4} m)
K.E, = 5.48 MeV
= 5.48 × 10^{6} × 1.6 × 10^{19} J
The correct answer is: 2
Calculate the typical K.E. expected of an α particle confined within a nucleus if its emitted energy is 10 MeV. (Answer in MeV)
Assume the nuclear potential to be square will of potential V = –10 MeV.
If the nuclear potential is a square well
From conservation of energy
The correct answer is: 20
Given that the nucleus density varies with r
where ρ_{0} = 0.14 nucleon/fm^{3}, R = 1.07 A^{1/3} and a = 0.54 fm.
Find the surface thickness i.e. the distance between which density drops from 0.9ρ_{0} to 0.1ρ_{0} (in fm upto two decimal places)
Dividing (2) by (1)
Surface thickness = r_{2} – r_{1} = a × 4.39
r_{2} – r_{1} = 0.54 × 4.39
= 2.37 fm
The correct answer is: 2.37
The binding energy of a heavy nucleus is about 7 MeV/nucleon, whereas the B.E of a medium weight nucleus is about 8 MeV/nucleon. Therefore the total K.E liberated when a heavy nucleus undergoes symmetric fission is (in MeV).
In case of symmetric fission the daughter nuclei have
A_{d} = A/2 each (daughter nuclei)
K.E. = diff. in B.E.
For a heavy nucleus A ~ 200
K.E. ~ 200 MeV
The correct answer is: 200
What should be the order of the energy of the electron beam which is to be used to explore the nuclear charge distribution (take the radius to be of the order of 10 fm) and also nucleon charge distribution (r = 0.8 fm)(in GeV)
For energy of electron,
λ ~ radius of the nucleus
= 10 fm = 10^{–14 }m
= 0.19 × 10^{10}eV ~ 2 × 10^{9}eV
~2 GeV
The correct answer is: 2
When α particles are directed onto atoms in a thin metal foil, some make very close collisions with the nuclei of the atom and are scattered at large angles. If an α particle with an initial kinetic energy of 5MeV happens to be scattered through an angle of 180°. Find the distance of closest approach to the scattering nucleus. (in fm)
The particle obeys the coulomb's law
(Helium) (Silver)
In SI units r = 2.9 × 10^{14}m
The correct answer is: 29
A sample of radioactive nuclei of a certain element can decay only by γ emission and β emission. If half life for γ emission is 24 minutes and that for γ emission is 36 minutes, the half life for the sample is how much? (in minutes)
In such cases, the decay constant is the sum of the different decay constants.
t_{1/2 }= 14.4 minutes
The correct answer is: 14.4
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