## When 92U235 undergoes fission. About 0.1% of the original mass is converted into energy. Then the amount of…..

Q. When 92U235 undergoes fission. About 0.1% of the original mass is converted into energy. Then the amount of 92U235 should undergo fission per day in a nuclear reactor so that it provides energy of 200 mega water electric power is

(a) 9.6 × 10-2 kg

(b) 4.8 × 10-2 kg

(c) 19.2 × 10-2 kg

(d) 1.2 × 10-2 kg

Ans: (c)

Sol: E = mc2

= 0.001 × (3×108)2

E = 10-3×9×1016 = 9×1013 J

Total mass of U required/sec

m= M×0.1% = M/1000

Power P = 200×106 J/s

200×106 = (M/1000)c2

M = 200×106 ×1000/c2

M = 0.22×10-5

Mass required per day = 0.22×10-5 ×24×60×60 = 19.2 ×10-2 kg

## The half life period of Pb210 is 22 years. If 2g of Pb210 is taken, then after 11 years the amount of Pb210 will be present is

Q. The half life period of Pb210 is 22 years. If 2g of Pb210 is taken, then after 11 years the amount of Pb210 will be present is

(a) 0.1414g

(b) 1.414g

(c) 2.828g

(d) 0.707g

Ans: (b)

Sol: N = N0(1/2)n , Where n = t/T1/2

N/N0 = (1/2)n

As , m ∝ N

m/m0 = (1/2)n

m/2 = (1/2)1/2 = 1/√2

m= 2/√2

m = √2 = 1.414

## If the activity of Ag^108 is 3 micro curie, the number of atoms present in it are

Q. If the activity of 108Ag is 3 micro curie, the number of atoms present in it are (λ = 0.005 sec-1)

(a) 2.2 × 107

(b) 2.2 × 106

(c) 2.2 × 105

(d) 2.2 × 104

Ans: (a)

Sol: 1 Curie = 3.7×1010 dps

1 micro Curie = 3.7×104 dps

Since , A = λ N

3 ×3.7×104 = 5×10-3 N

N = (3 ×3.7×104)/(5×10-3)

= 2.2 × 107

## A nucleus with mass number 220 initially at rest emits an α-particle. If the Q value of the reaction is 5.5 MeV then….

Q. A nucleus with mass number 220 initially at rest emits an α-particle. If the Q value of the reaction is 5.5 MeV then the kinetic energy of the α-particle is (in MeV)

(a) 4.4

(b) 5.4

(c) 5.6

(d) 6.5

Ans: (b)

Sol: E = P2/2m

As, P = constant

mE = constant

## A nucleus splits into two nuclear parts having radii in the ratio 1 : 2. Their velocities are in the ratio

Q. A nucleus splits into two nuclear parts having radii in the ratio 1 : 2. Their velocities are in the ratio

(a) 8 : 1

(b) 6 : 1

(c) 4 : 1

(d) 2 : 1

Ans: (a)

Sol: Applying conservation of momentum ,

m1v1 = m2v2 $\displaystyle \rho \frac{4}{3}\pi r_1^3 v_1 = \rho \frac{4}{3}\pi r_2^3 v_2$ $\displaystyle r_1^3 v_1 = r_2^3 v_2$ $\displaystyle \frac{v_1}{v_2} = (\frac{r_2}{r_1})^3 = (2)^3$ $\displaystyle \frac{v_1}{v_2} = 8$