A 1-0 MeV neutron emitted in a fusion loses one half of its kinetic energy in each collision with a moderator molecule- Number of collision it must be to reach thermal energy will be left-Take- dfrac -3-2-k-B-T-0-040 eVright- 512562525

Question

Toppr Admin · Accepted Answer

-xA0-Let -quot-n-quot- be required number-xA0-xA0-of collision-xA0-xA0-Given-xA0-32kBT-0-04-0-04-n-1n-25

A $1.0 M e V$ neutron emitted in a fusion loses one half of its kinetic energy in each collision with a moderator molecule. Number of collision it must be more to reach thermal energy will be $(T a k e, \frac{3}{2} k_{B} T = 0.040 e V)$
$5$
$125$
$625$
$25$

$\begin{matrix} Let "n" be required number of collision Given \frac{3}{2} k_{B} T = 0.04 (0.04) n = 1 n = 25 \end{matrix}$

A 1.0 MeV neutron emitted in a fusion loses one half of its kinetic energy in each collision with a moderator molecule. Number of collision it must be more to reach thermal energy will be (Take,32kBT=0.040 eV) 512562525

Let "n" be required number of collision Given 32kBT=0.04(0.04)n=1n=25

A $1.0 M e V$ neutron emitted in a fusion loses one half of its kinetic energy in each collision with a moderator molecule. Number of collision it must be more to reach thermal energy will be $(T a k e, \frac{3}{2} k_{B} T = 0.040 e V)$
$5$
$125$
$625$
$25$

$\begin{matrix} Let "n" be required number of collision Given \frac{3}{2} k_{B} T = 0.04 (0.04) n = 1 n = 25 \end{matrix}$