Calculate the binding energy per nucleon of Kr 36 84 whose atomic mass is 83 . 913 u . (Mass of neutron is that of H atom is 1 . 0078 u )

Atomic mass of Kr 36 84 ( m ) = 83 . 913 u Mass of neutron ( m n ) = 1 . 0087 u Mass of H atom ( m H ) = 1 . 0078 u Total mass = Z . m H + ( A - Z ) M n = ( 36 × 1 . 0078 ) + ( 48 × 1 . 0087 ) = 84 . 6984 u Mass defect ( ∆ m ) = Total mass - m = 84 . 6984 - 83 . 931 ∆ m = 0 . 7674 u Total binding energy = ∆ m × 931 . 4 Mev = 0 . 7674 × 931 . 4 = 714 . 756 Mev Binding energy per nucleon = Binding energy Number of nucleons = 714 . 756 84 = 8 . 5 Mev

Calculate the binding energy per nucleon of Kr 36 84 whose atomic mass is 83 . 913 u . (Mass of neutron is that of H atom is 1 . 0078 u )

Atomic mass of Kr 36 84 ( m ) = 83 . 913 u Mass of neutron ( m n ) = 1 . 0087 u Mass of H atom ( m H ) = 1 . 0078 u Total mass = Z . m H + ( A - Z ) M n = ( 36 × 1 . 0078 ) + ( 48 × 1 . 0087 ) = 84 . 6984 u Mass defect ( ∆ m ) = Total mass - m = 84 . 6984 - 83 . 931 ∆ m = 0 . 7674 u Total binding energy = ∆ m × 931 . 4 Mev = 0 . 7674 × 931 . 4 = 714 . 756 Mev Binding energy per nucleon = Binding energy Number of nucleons = 714 . 756 84 = 8 . 5 Mev

Calculate the energy in Mev released in the nuclear reaction Ir 77 174 → Re 75 170 + He 2 4 . Atomic masses : Ir = 173 . 97 u , Re = 169 . 96 u and He = 4 . 0026 u

Ir 77 174 → Re 75 170 + He 2 4 Mass of Ir = 173 . 97 u Mass of Re = 169 . 96 u Mass of He = 4 . 0026 u Mass defect = ( mass of Ir ) - [ ( mass of Re ) + ( mass of He ) ] = 173 . 997 - 173 . 96 = 0 . 0074 u The energy released = ∆ m × 931 . 4 Mev = 0 . 0074 × 931 . 4 = 6 . 89 Mev

How many α -particles are emitted by 0 . 1 g of Ra 226 in one year?

Activity = Number of α -particles emitted. - dN dt = λN ∵ Half life of Ra = 1600 years λ = 0 . 693 1600 year - 1 ( ∵ λ = 0 . 693 t 1 / 2 ) N = 0 . 1 g × 6 . 022 × 10 23 mol - 1 226 g mol - 1 = 2 . 66 × 10 20 Therefore, number of particles emitted per year = λN = 0 . 693 1600 × 2 . 66 × 10 20 = 1 . 152 × 10 17 .

It has been found that the Sun's mass loss is 4 . 34 × 10 9 kg per second. How much energy per second would be radiated into space by the Sun?

To calculate the energy radiated, we use the mass-energy relation, E = mc 2 Here, m is the mass and c is the speed of light. Given, m = 4 . 34 × 10 9 kg The value of c is 3 × 10 8 ms - 1 Substituting these values in the mass-energy relation, we get E = 4 . 34 × 10 9 × ( 3 × 10 8 ) 2 = 4 . 34 × 9 × 10 25 = 39 . 06 × 10 25 = 3 . 9 × 10 26 J s - 1