Calculate the binding energy per nucleon for Ne 10 20 . Given that m H = 1 . 007825 u , m n = 1 . 008665 u and mass of Ne 10 20 atom 19 . 992440 u .

For N 10 20 e : mass defect, ∆ m = 10 m p + 20 - 10 m n + M N e ∆ m = 10 × 1 . 007825 + 10 * 1 . 008665 - 19 . 9924 = 0 . 1725 a m u Binding energy per nucleon B E = ∆ m c 2 A = 0 . 1725 c 2 20 = 0 . 0086 c 2 a m u = 0 . 0086 × 931 . 5 = 8 . 03 M e V a s 1 a m u = 931 . 5 M e V / c 2 ...

State the radio-active decay law. Deduce the relation N = N o e - 2 λt , where the symbols have their usual meanings

The radioactive decay law states that the probability per unit time that a nucleus will decay is constant, independent of time. The radioactive decay of a certain number of atoms (mass) is exponential in time. We know by radioactive decay law d N d T ∝ N o r d N d T = - λ N . Where λ is the constant of proportionally called radioactive decay constant. Let N ∘ be the number of nuclei present at time t=0 ,and N the number of nuclei present at time t. ∫ N ∘ N d N N = - ∫ o t λ d t = - λ ∫ o t d t log e N - log e N o = - λ t . log e ( N N o ) = - λ t N N o = e - λ t , N = N o e - λ t .

Two nuclei have mass numbers in the ratio 1 : 8 . What is the ratio of their nuclear radii?

A 1 = 1 , A 2 = 8 R 1 R 2 = A 1 A 2 3 Therefore by putting the values,ratio in radii will be 1 : 2 we get, 1 : 2 ...

Explain the process of release of energy in a nuclear reactor. Draw a schematic diagram of a nuclear reactor and write the function of each part.

Nuclear Reactor With the help of nuclear reactors, power is generated by changing the energy obtained from nuclear into energy. As we have already mentioned the controlled obtain reactions are used in nuclear reactors. Here we will discuss nuclear reactor based on fission of 235 U in Russia (Ukraine in present, such incident happened in Chernobyl nuclear plant, whose result can till be seen in the local people. India Atomic Energy Programme is in accordance with the international systems in view of safety. Fuel:The fissionable material used in the reactor is called as fuel. The commonly used fuels are Uranium, Plutonium or Thorium. Moderators: Neutrons of a fairly low speed should be used to have controlled chain reaction. To slow down the speed fast moving neutrons produced during the fission process, moderators are used. Moderator reduces the speed of the neutron by absorbing its energy but not absorb neutron. Graphite, Heavy water and Beryllium are common moderators. Control Rods:These rods absorb neutrons and stop the chain reaction to proceed further. These are made up of steel containing a high percentage of material like cadmium or boron which can absorb neutrons. Shielding: Shielding prevents radiations to reach outside the reactor and protects the personnel associated with the operations. Lead blocks and concrete enclosure that is strong enough of several meters thickness are used for shielding. Coolant: The coolant is where heat-exchange process occurs. Heat is absorbed by the coolant that is produced in the reactor. Turbines:Steam produced in the boiler is now passes to a turbine and the force of the steam jet causes the turbine to rotate. Heat energy (steam) is converted to mechanical energy (moving turbine). Generator:The generator consists of coils that change the mechanical energy into electric energy. ..

Define the terms “half-life period” and “decay constant” of a radioactive sample. Derive the relation between them.

Half-life period: The half-life period of an element is defined as the time in which the number of radioactive nuclei decay to half of its initial value. Decay constant: The decay constant of a radioactive element is defined as the reciprocal of time in which the number of undecayed nuclei of that radioactive element fails to 1 e times of its initial time. The relation between half-life and decay constant-The radioactive decay equation is N = N o e - λ t . . . . . . ( i ) when t = T , N = N o 2 ∴ N o 2 = N o e - λ T o r e - λ T = 1 2 . . . . . . . ( i i ) Taking log of both sides - λ T log e = log e 1 - log e 2 o r λ T = log e 2 ∴ T = log e 2 λ = 2 . 3026 log 10 2 λ = 2 . 3026 × 0 . 3010 λ o r T = 0 . 6931 λ . ..

Calculate the binding energy per nucleon for Ne 10 20 . Given that m H = 1 . 007825 u , m n = 1 . 008665 u and mass of Ne 10 20 atom 19 . 992440 u .

For N 10 20 e : mass defect, ∆ m = 10 m p + 20 - 10 m n + M N e ∆ m = 10 × 1 . 007825 + 10 * 1 . 008665 - 19 . 9924 = 0 . 1725 a m u Binding energy per nucleon B E = ∆ m c 2 A = 0 . 1725 c 2 20 = 0 . 0086 c 2 a m u = 0 . 0086 × 931 . 5 = 8 . 03 M e V a s 1 a m u = 931 . 5 M e V / c 2 ...

Mass of an electron is 9 . 11 × 10 - 31 kg . Calculate mass-energy of electron in Joule as well as electron volts. Given that speed of light in vacuum c = 3 . 0 × 10 8 ms - 1 and 1 eV = 1 . 60 × 10 - 19 J .

Mass-energy of electron = m c 2 . E = 9 . 11 × 10 - 31 × 3 × 10 8 2 = 8 . 2 × 10 - 14 J Given, 1 eV = 1 . 60 × 10 - 19 J , So, 1 J = 1 1 . 60 × 10 - 19 eV . Thus, electron volts equivalent is = 8 . 2 × 10 - 14 1 . 6 × 10 19 = 51 . 25 × 10 - 4 eV .