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Nuclear Physics-Multiple choice questions

1. A hospital patient receives an x-ray dose of 0.5 m rad. What is the dose equivalent in sieverts?

a. 0.5 m Sv
b. .005 m Sv
c. .05 m Sv
d. 5 m Sv

2. Determine the identity of the nucleus X in the reaction 13C6 + n -----> X + gamma

a. 13C6
b. 14C6
c. 14N7
d. 14N6

3. Identify the particle Y in the reaction 2H1 + Y ------> 4He2 + n

a. 1H1
b. 2H1
c. 3H1
d. 4H1

4. What is the kinetic energy in MeV of a room-temperature thermal neutron? By "thermal neutron" we mean one with kinetic energy kT. take room temperature to be 300 K.

a. 2.59 x 10^(-8) MeV
b. 2.59 x 10^(-6) MeV
c. 2.59 x 10^(-5) MeV
d. 2.59 x 10^(-3) MeV

5. How much mass is lost in the fission of the nuclear fuel in a 1000 MW power plant in one year? (assume that the power plant is 33% efficient)

a. 2.10 kg
b. 3.15 kg
c. .35 kg
d. 1.05 kg

6. Calculate the energy released in the reaction 3He2 + 3He2 ----> 4He2 + 1H1 + 1H1

a. 12.86 MeV
b. 6.42 MeV
c. 3.12 MeV
d. 18.51 MeV

Solution This solution is FREE courtesy of BrainMass!

1. A hospital patient receives an x-ray dose of 0.5 m rad. What is the dose equivalent in sieverts?

a. 0.5 m Sv
b. .005 m Sv
c. .05 m Sv
d. 5 m Sv

Answer: b. .005 m Sv

rem = rad x Q.
X rays and gamma rays have a Q about 1
Therefore 1 rem = 1 rad as Q=1
100 rem = 1 Sv
Therefore 0.5m rad= 0.5 m rem = 0.5 m /100 Sv= .005 m Sv

2. Determine the identity of the nucleus X in the reaction 13C6 + n -----> X + gamma

a. 13C6
b. 14C6
c. 14N7
d. 14N6

Answer: b. 14C6

n has a charge 0 and mass 1
gamma rays are radiation: no charge and mass
Therefore
X will have a charge = 6+0=6
Mass= 13+1=14
Therefore X = 14C6

3. Identify the particle Y in the reaction 2H1 + Y ------> 4He2 + n

a. 1H1
b. 2H1
c. 3H1
d. 4H1

Answer: c. 3H1

Let the atomic no of Y be x and Mass numebr be y
n has a charge of 0 and mass of 1

2+ y = 4+1
or y= 3

1+x= 2+0
or x=1

Therefore Y = 3H1

4. What is the kinetic energy in MeV of a room-temperature thermal neutron? By "thermal neutron" we mean one with kinetic energy kT. take room temperature to be 300 K.

a. 2.59 x 10^(-8) MeV
b. 2.59 x 10^(-6) MeV
c. 2.59 x 10^(-5) MeV
d. 2.59 x 10^(-3) MeV

Answer: a. 2.59 x 10^(-8) MeV

k = 1.38E-23 J/'K
T= 300 K
Therefore kT= 4.14E-21 J

1 MeV= 1.60E-13 J

Therefore kT = 4.14E-21 J= 2.59E-08 MeV

5. How much mass is lost in the fission of the nuclear fuel in a 1000 MW power plant in one year? (assume that the power plant is 33% efficient)

a. 2.10 kg
b. 3.15 kg
c. .35 kg
d. 1.05 kg

Answer: d. 1.05 kg

1 MW= 1.00E+06 J/s
1 year= 3.15E+07 s

Therefore total energy produced= 3.15E+16 Joules

E=mc^2
c= 3.00E+08 m/s
E= 3.15E+16 Joules

Therefore m= 0.35 Kg

Dividing this value by efficiency factor we get the actual mass required

Efficiency= 33%

Therefore mass= 1.06 Kg

6. Calculate the energy released in the reaction 3He2 + 3He2 ----> 4He2 + 1H1 + 1H1

a. 12.86 MeV
b. 6.42 MeV
c. 3.12 MeV
d. 18.51 MeV

Answer: a. 12.86 MeV

mass of 4 He2 = 4.0026 u
mass of 3 He2 = 3.01603 u
mass of 1 H1 = 1.007825 u

Total mass of 3He2 + 3He2= 6.03206 u
Total mass of 4He2 + 1H1 + 1H1= 6.01825 u

difference= 0.01381 u
conversion between rest mass and energy is 931.494 MeV/u

Therefore energy= 12.86 MeV

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