Home
Class 12
PHYSICS
A nuclear reactor using .^(235)U generat...

A nuclear reactor using `.^(235)U` generates `250 MW` of electric power. The efficiency of the reactor (i.e., efficiency of conversion of thermal energy into electrical energy) is `25%`. What is the amount of `.^(235)U` used in the reactor per year? The thermal energy released per fission of `.^(235)U` is `200 MeV`.

Promotional Banner

Similar Questions

Explore conceptually related problems

If 5g of .^(235)U is completely destroyed in a reactor, the energy released would be

Calculate the amount of ""_(92)U^(235) required to release energy of 1 kWh. Given energy released during fission of one atom is 200 MeV.

A uranium rector develops thermal energy at a rate of 300 MW. Calculate the amount of ^235U being consumed every second .Average energy released per fission is 200 MeV.

The amount of U^(235) to be fissioned to operate 10 kW nuclear reactor is

Calculate the energy released by fission from 2 g of .^(235)._(92)U in kWh . Given that the energy released per fission is 200 MeV .

How much energy is released when 2 mole of U^(235) is fission :

Suppose India has a target of producing by 2020 A.D., 2xx10^(5)MW of electric power, ten percent of which is to be obtained form nuclear power plants. Suppose we are given that on an average, the efficiency of utilization (i.e., conservation to electrical energy) of thermal energy produced in a reactor is 25% . How much amount of fissionable uranium will our country need per year? Take the heat energy per fission U^(235) to be about 200MeV.

Suppose India has a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an avedrage, the efficiency of utilization(i.e conversion to electric energy) of thermal energy produced in a reactor was 25% . How much amount of fissionable uranium would our country need per year by 2020 ? Take the heat energy per fission of .^(235)U to be about 200 MeV .

Calculate the energy released by the fission 1 g of .^(235)U in joule, given that the energy released per fission is 200 MeV . (Avogadro's number =6.023xx10^(23))