The question seems to be a bit jumbled, but I understand that you're asking how many fissions per second are required to release 2 kW of power, given that each fission of a 92U235 nucleus releases 200 MeV of energy. Here's how you can calculate it:

1. First, convert the energy released per fission from MeV to Joules. 1 MeV = 1.60218 x 10^-13 Joules. So, 200 MeV = 200 x 1.60218 x 10^-13 Joules = 3.20436 x 10^-11 Joules.

2. Next, convert the power from kW to Joules per second. 1 kW = 1000 Joules per second. So, 2 kW = 2000 Joules per second.

3. Now, divide the total power by the energy released per fission to find the number of fissions per second. So, 2000 Joules per second / 3.20436 x 10^-11 Joules per fission = 6.24 x 10^14 fissions per second.

So, approximately 6.24 x 10^14 fissions per second are required to release 2 kW of power.

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The question seems to be a bit jumbled, but I understand that you're asking how many fissions per second are required to release 2 kW of power, given that each fission of a 92U235 nucleus releases 200 MeV of energy. Here's how you can calculate it:

1. First, convert the energy released per fission from MeV to Joules. 1 MeV = 1.60218 x 10^-13 Joules. So, 200 MeV = 200 x 1.60218 x 10^-13 Joules = 3.20436 x 10^-11 Joules.

2. Next, convert the power from kW to Joules per second. 1 kW = 1000 Joules per second. So, 2 kW = 2000 Joules per second.

3. Now, divide the total power by the energy released per fission to find the number of fissions per second. So, 2000 Joules per second / 3.20436 x 10^-11 Joules per fission = 6.24 x 10^14 fissions per second.

So, approximately 6.24 x 10^14 fissions per second are required to release 2 kW of power.

Knowee AI · Accepted Answer

The question seems to be a bit jumbled, but I understand that you're asking how many fissions per second are required to release 2 kW of power, given that each fission of a 92U235 nucleus releases 200 MeV of energy. Here's how you can calculate it:

1. First, convert the energy released per fission from MeV to Joules. 1 MeV = 1.60218 x 10^-13 Joules. So, 200 MeV = 200 x 1.60218 x 10^-13 Joules = 3.20436 x 10^-11 Joules.

2. Next, convert the power from kW to Joules per second. 1 kW = 1000 Joules per second. So, 2 kW = 2000 Joules per second.

3. Now, divide the total power by the energy released per fission to find the number of fissions per second. So, 2000 Joules per second / 3.20436 x 10^-11 Joules per fission = 6.24 x 10^14 fissions per second.

So, approximately 6.24 x 10^14 fissions per second are required to release 2 kW of power.

Assuming that 200 MeV of energy is released per fission of 92U235 nucleus. Find the number of fission per second required to release 2 kW power3.125 × 10136.25 × 10133.125 × 10156.25 × 1014

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