Ask question

Ask question

All categories

3 years ago

8

In a liquid metal fast breeder reactor, no neutron moderation is desired and sodium is used as a coolant to minimize fission-neu

tron thermalization. How many elastic scatters with sodium, on the average, would it take for 2-MeV neutrons to reach an average thermal energy of 0.025 eV?

1 answer:

ololo11 [35]3 years ago

5 0

Answer:

219 scatterings

Explanation:

Given that:

The Coolant used In the liquid metal fast breed reactor = Sodium

The atomic weight (A) of sodium = 23

The initial energy $E_{i}$ = 2 - MeV

The final energy $E_{f}$ = 0.025 eV (thermal energy)

The number of elastic neutron scatterings (n) needed to reach the given average thermal energy can be computed as:

$n = \dfrac{log \bigg(\dfrac{E_f}{E_i} \bigg)}{log \bigg [ \dfrac{A^2+1}{(A+1)^2} \bigg]}$

$n = \dfrac{log \bigg(\dfrac{0.025}{2 \times 10^6} \bigg)}{log \bigg [ \dfrac{23^2+1}{(23+1)^2} \bigg]}$

$n = \dfrac{log \bigg(1.25\times 10^{-8} \bigg)}{log \bigg [ 0.92014\bigg]}$

$n = 218.643$

n ≅ 219 scatterings

You might be interested in

A nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the

Sladkaya [172]

Answer:

(a). The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$ .

(b). The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$ .

Explanation:

Given that,

Radius of sphere R=2.81 cm

Charge = +2.35 fC

Potential at center of sphere

$V = 0$

(a). We need to calculate the potential at a distance r = 1.60 cm

Using formula of potential difference

$V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr$

$V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr$

$V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}$

$V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})$

$V_{r}=-0.00012190\ V$

$V_{r}=-1.219\times10^{-4}\ V$

The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$ .

(b). We need to calculate the potential at a distance r = R

Using formula of potential difference

$V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}$

$V_{R}=-0.0003759\ V$

$V_{R}=-3.759\times10^{-4}\ V$

The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$ .

Hence, This is the required solution.

5 0

3 years ago

Please help 9.2.1 project in science just ned an example

olga55 [171]

Answer:

Give me what kind of example you need please so I can help you. Put it in the comments.

Explanation:

3 0

3 years ago

What method of heat transfer does heat energy use to reach earth from the sun?

Alchen [17]

Through radiation of the sun's rays.

4 0

4 years ago

Read 2 more answers

What is transferred by a force moving an object through a distance?

Komok [63]

<span>When you apply force to move an object at a distance, you are applying work. And work is energy in transit. The answer is letter D. For example, you see a cart at a distance. You observe that it is not moving. You want to transfer it to your backyard. You apply force to the cart and observed that the cart is not at the same position as it was before. You are applying work to the cart by transferring your energy to it.</span>

7 0

3 years ago

Read 2 more answers

Sand falls from an overhead bin and accumulates in a conical pile with a radius that is always threethree times its height. Supp

hammer [34]

Answer:

159241.048 cm³/s

Explanation:

r = Radius = 3×height = 3h

h = height = 16 cm

Height of the pile increases at a rate = $\frac{dh}{dt}=22\ cm/s$

$\text{Volume of cone}=\frac{1}{3}\pi r^2h\\\Rightarrow V=\frac{1}{3}\pi (3h)^2h\\\Rightarrow V=3\pi h^3$

Differentiating with respect to time

$\frac{dv}{dt}=9\pi h^2\frac{dh}{dt}\\\Rightarrow \frac{dv}{dt}=9\pi 16^2\times 22\\\Rightarrow \frac{dv}{dt}=159241.048\ cm^3/s$

∴ Rate is the sand leaving the bin at that instant is 159241.048 cm³/s

5 0

3 years ago