Tag: nuclear physics

Quizzes (6) Questions (198)

Questions Related to nuclear physics

Multiple choice chemistry nuclear physics nuclear reactor the nuclear power station isotopes and nuclear chemistry

Moderator is a substance that ..................... the neutrons in a nuclear reactor.

  1. Slow down

  2. Activates

  3. Both (a) and (b) above

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A moderator is a material used in nuclear reactors to reduce the speed of fast neutrons, thereby increasing the probability of them causing further fission in U-235 nuclei.

Multiple choice chemistry nuclear physics nuclear reactor the nuclear power station isotopes and nuclear chemistry

The amount of $U^{235}$ to be fissioned, to operate a $10\ kW$ nuclear reactor is (Approximately)

  1. $1.2 \times 10 ^ { - 5 }\ \mathrm { gm } / \mathrm { s }$
  2. $1.2 \times 10 ^ { - 7 }\ \mathrm { gm } / \mathrm { s }$
  3. $1.2 \times 10 ^ { - 9 }\ \mathrm { gm } / \mathrm { s }$
  4. $1.2 \times 10 ^ { - 11 }\ \mathrm { gm } / \mathrm { s }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the relation P = (n * E) / t, where P is power, n is number of fissions, and E is energy per fission. 10 kW = 10,000 J/s. E = 200 MeV = 200 * 1.6 * 10^-13 J. Calculating the mass of U-235 required leads to approximately 1.2 * 10^-11 g/s.

Multiple choice chemistry nuclear physics nuclear reactor the nuclear power station isotopes and nuclear chemistry

The amount of $U ^ { 235 }$ in kg which is to be used per hour in a nuclear reactor of capacity $100$ ($E = 200 MeV/fission$)

  1. $0.45 \times 10^{-5}$
  2. $4.5 \times 10^{-5}$
  3. $4.5 \times 10^{5}$
  4. $45 \times 10^{-5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power = 100 MW = 10^8 J/s. Energy per fission = 200 MeV = 3.2 * 10^-11 J. Fissions per second = 10^8 / (3.2 * 10^-11) = 3.125 * 10^18. Mass per second = (3.125 * 10^18 * 235) / (6.022 * 10^23) = 1.22 * 10^-3 g/s. Multiplying by 3600 seconds gives the hourly rate.

Multiple choice chemistry nuclear physics nuclear reactor the nuclear power station isotopes and nuclear chemistry

Cadmium rods are used as moderators in a nuclear reactor.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Cadmium rods are used in a nuclear reactor for regulating the power level of the reactor. Cadmium rods are used in nuclear reactors to control the fission rate of radioactive material used. It is capable of absorbing many neutrons without fissioning themselves.
Heavy water, graphite etc. are used as moderator
Multiple choice chemistry nuclear physics nuclear reactor the nuclear power station isotopes and nuclear chemistry

An atomic power nuclear reactor can deliver $300$ MW. The energy released due to fission of each nucleus of uranium atom $U^{235} $ is $170$ MeV. The number of uranium atoms fissioned per hour will be 

  1. $4\times 10^{22}$
  2. $30\times 10^{22}$
  3. $10\times 10^{20}$
  4. $5\times 10^{15}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total energy released per second is 300 MW, which equals 300 x 10^6 J/s. Dividing this by the energy per fission in joules gives the total number of fissions per second, and multiplying by 3600 yields the number of fissions per hour, approximately 4 x 10^22.

Multiple choice chemistry nuclear physics nuclear reactor the nuclear power station isotopes and nuclear chemistry

The fission of one uranium nucleus releases energy of amount x joules. Find the number of fissions required to produced energy at the rate of v MW for t hours a nuclear power plant.

  1. $ \dfrac {yt}{x} $
  2. $ \dfrac{x}{yt} $
  3. $ 3.6 \times 10^9 \dfrac {yt}{x} $
  4. $ 3.6 \times 10^9 \dfrac {x}{yt} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total energy required = Power (v MW) * time (t hours). Total energy = v * 10^6 * t * 3600 Joules. Number of fissions = Total energy / energy per fission (x). This results in the expression involving v, t, and x.