200 MeV of energy can be obtained per fission. In a reactor generating 1000 kW, the number of nuclei under going the fission per second is 3.125 × 10n. Find n.

Question

200 MeV of energy can be obtained per fission. In a reactor generating 1000 kW, the number of nuclei under going the fission per second is 3.125 × 10n. Find n. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

P=1000 kW P=( text Energy / text Time ) P=( text Energy released per fission × text Number of fission / text Time ) 1000 × 103=(3.2 × 10-11 × text Number of fission / text Time ) Number of fission per second =3.125 × 1016

Q. $200 M e V$ of energy can be obtained per fission. In a reactor generating $1000 kW$ , the number of nuclei under going the fission per second is $3.125 \times 1 0^{n}$ . Find $n$ .

$P = 1000 kW$
$P = \frac{Energy}{Time}$
$P = \frac{Energy released per fission \times Number of fission}{Time}$
$1000 \times 1 0^{3} = \frac{3.2 \times 1 0 ^{- 11} \times Number of fission}{Time}$
Number of fission per second $= 3.125 \times 1 0^{16}$

Q. 200MeV of energy can be obtained per fission. In a reactor generating 1000kW, the number of nuclei under going the fission per second is 3.125×10n. Find n.

P=1000kW P= Time Energy ​ P= Time Energy released per fission × Number of fission ​ 1000×103= Time 3.2×10−11× Number of fission ​ Number of fission per second =3.125×1016

Q. $200 M e V$ of energy can be obtained per fission. In a reactor generating $1000 kW$ , the number of nuclei under going the fission per second is $3.125 \times 1 0^{n}$ . Find $n$ .

$P = 1000 kW$
$P = \frac{Energy}{Time}$
$P = \frac{Energy released per fission \times Number of fission}{Time}$
$1000 \times 1 0^{3} = \frac{3.2 \times 1 0 ^{- 11} \times Number of fission}{Time}$
Number of fission per second $= 3.125 \times 1 0^{16}$