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Physics Part-II

Nuclei

Suppose India had a target of producing by 2020 AD, 200,000 MW

Physics Part-II

Class 12

NCERT

Chapter 1: Ray Optics and Optical Instruments

41 questions

Chapter 2: Wave Optics

30 questions

Chapter 3: Dual Nature of Radiation and Matter

44 questions

Chapter 4: Atoms

23 questions

Chapter 5: Nuclei

38 questions

Calculate and compare the energy released by (a) Fusion of 1.0 kg of hydrogen deep within Sun (b) The fission of 1.0 kg of

^{235} U

in a fission reactor

Suppose India had 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 average, 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.

Given the mass of iron nucleus as

55.85 u

and

A = 56

, find the nuclear density?

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Chapter 6: Semiconductor Electronics: Materials, Devices and Simple Circuits

25 questions

Question

Medium

Solving time: 5 mins

Suppose India had 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 average, 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.

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Updated on: Sep 25, 2022

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Physics Part-II (NCERT)

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^{235} U

to be about 200 MeV.

15 mins ago

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Text solutionVerified

The amount of electric power to be generated,

P = 2 \times 1 0^{5}

10

% of this power has to be obtained from nuclear power plants.

Amount of nuclear power:

P_{1} = \frac{10}{100} \times 2 \times 1 0^{5}

= 2 \times 1 0^{10} \times 3600 \times 24 \times 365 \times J / ye a r

Heat energy released per fission of a Uranium nucleus,

E = 200 M e V

Efficiency of reactor

= 25%

Hence, the amount of energy converted into electrical energy per fission is calculated as:

\frac{25}{100} \times 200 = 50 M e V = 8 \times 1 0^{- 12} J

Number of atoms required for fission per year:

\frac{2 \times 1 0 ^{10} \times 3600 \times 24 \times 365}{8 \times 1 0 ^{- 12}}

1

mole, i.e.

235 g

of Uranium contains

6.023 \times 1 0^{23}

atoms

Mass of

6.023 \times 1 0^{23}

atoms of Uranium is

= 235 g

Mass of

78840 \times 1 0^{24}

atoms of Uranium

= 3.076 \times 1 0^{4} k g

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Practice more questions from Physics Part-II (NCERT)

Calculate and compare the energy released by (a) Fusion of 1.0 kg of hydrogen deep within Sun (b) The fission of 1.0 kg of

^{235} U

in a fission reactor

^{235} U

to be about 200 MeV.

Given the mass of iron nucleus as

55.85 u

and

A = 56

, find the nuclear density?

View all

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Question Text	Suppose India had 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 average, 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.
Updated On	Sep 25, 2022
Topic	Nuclei
Subject	Physics
Class	Class 12
Answer Type	Text solution:1 Video solution: 5
Upvotes	543
Avg. Video Duration	17 min

Text solutionVerified

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