Question
A heavy nucleus breaks into comparatively lighter nuclei, which are more stable compared to the original heavy nucleus. When a heavy nucleus like uranium is bombarded by slow moving neutrons, it splits into two parts, releasing large amount of energy. The typical fission reaction of $_{92}\text{U}^{235}$.
$_{92}\text{U}^{235}+\ _0\text{n}^1\rightarrow\ _{56}\text{Ba}^{141}+\ _{36}\text{kr}^{92}+\ 3_0\text{n}^1+\ 200\text{ MeV}$
The fission of $_{92}\text{U}^{235}$ approximately released 200 MeV of energy.
  1. If 200 MeV energy is released in the fission of a single nucleus of $_{92}^{235}\text{U}$, the fissions which are required to produce a power of 1kW is.
  1. $3.125 \times 10^{13}$
  2. $1.52 \times 10^6$
  3. $3.125 \times 10^{12}$
  4. $3.125 \times 10^{14}$
  1. The release in energy in nuclear fission is consistent with the fact that uranium has
  1. More mass per nucleon than either of the two fragments.
  2. More mass per nucleon as the two fragment.
  3. Exactly the same mass per nucleon as the two fragments.
  4. Less mass per nucleon than either of two fragments.
  1. When $_{92}\text{U}^{235}$ undergoes fission, about 0.1% of the original mass is converted into energy. The energy released when 1kg of $_{92}\text{U}^{235}$ undergoes fission is.
  1. $9 \times 10^{11}J$
  2. $9 \times 10^{13}J$
  3. $9 \times 10^{15}J$
  4. $9 \times 10^{18}J$
  1. A nuclear fission is said to be critical when multiplication factor or K.
  1. $K = 1$
  2. $K > 1$
  3. $K < 1$
  4. $K = 0$
  1. Einstein's mass-energy conversion relation $E = mc^2$ is illustrated by.
  1. Nuclear fission
  2. $\beta-\text{decay}$
  3. Rocket propulsion
  4. Steam engine

Answer

  1. (a) $3.125 \times 10^{13}$
Explanation:
Let the number of fissions per second be n. Energy released per second
$= n \times 200MeV = n \times 200 \times 1.6 \times 10^{-13}J$
Energy required per second = power × time
$= 1kW × Is = 1000J$
$n \times 200 \times 1.6 \times 10^{-13} = 1000$ or
$\text{n}=\frac{1000}{3.2\times10^{-11}}$
$=\frac{10}{3.2}\times10^{13}=3.125\times10^{13}$
  1. (a) More mass per nucleon than either of the two fragments.
  2. (b) $9 \times 10^{13}J$
Explanation:
As only 0.1% of the original mass is converted into energy, hence out of 1kg mass 1g is converted into energy.
Energy released during fission $\text{E}=\triangle\text{mc}^2$
$= 1g \times (3 \times 10^8 m s^{-1})^2 = 10^{-3} \times 9 \times 10^{16}J = 9 \times 10^{13}J$
  1. (a) K = 1
  2. (a) Nuclear fission

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