Suppose the target in a laser fusion reactor is a sphere of...

Suppose the target in a laser fusion reactor is a sphere of solid hydrogen that has a diameter of $1.50 \times 10^{-4} \mathrm{m}$ and a density of $0.200 \mathrm{g} / \mathrm{cm}^{3} .$ Assume half of the nuclei are $^{2} \mathrm{H}$ and half are $^{3} \mathrm{H} .$ (a) If 1.00$\%$ of a $200-\mathrm{kJ}$ laser pulse is delivered to this sphere, what temperature does the sphere reach? (b) If all the hydrogen "burns" according to the D-T reaction, how many joules of energy are released?

Key Concepts

Energy Deposition in Laser Systems

This concept involves understanding how a fraction of a laser's energy is transferred to a target. It encompasses calculating the energy actually absorbed by the material from a high-powered source, an essential step in controlled fusion experiments where precise energy delivery is critical for achieving the desired reaction conditions.

Thermal Energy and Temperature Calculation

This concept includes using the energy absorbed by a material to calculate its temperature rise. It involves determining the amount of heat energy deposited and relating it to the temperature increase via the material’s specific heat capacity or by idealized assumptions about energy distribution in the system.

Density, Volume, and Mass Relationships

This topic covers the geometric and physical relations used to determine the mass of a material from its density and volume. In the context of spherical targets, it requires calculating the sphere’s volume and then applying the density to find the mass, which is crucial for both thermal and nuclear energy calculations.

Nuclear Fusion Reactions

Nuclear fusion reactions, such as the deuterium–tritium (D-T) reaction, are a key concept where light nuclei combine to form heavier nuclei, releasing substantial energy in the process. Understanding the specific fusion reaction, including the type of fuel used and the energy released per reaction, is central to evaluating the potential energy output in fusion reactors.

Fusion Energy Yield Calculation

This concept involves computing the total energy released from a given mass of fusion fuel once the reactions occur. It requires knowledge of the energy per reaction event, the proportion of different isotopes present, and the total number of reaction events, thereby connecting nuclear reaction theory to practical energy generation estimates.

Transcript

00:01 Hello friends this is a problem based on laser fusion reactor so it is based on laser fusion reactor given there is a solid hydrogen sphere of diameter so diameter of solid hydrogen sphere are equal to 1 .5 into 10 to the power minus 4 meter and its density is 0 .2 gram per centimeter cube that is 200 kg per meter cube it having half of 1 h2 nuclei and remaining half 1 h3 nuclei first part we have to find the temperature of spare if 1 % of 200 kilojoules laser pulse delivered to the sphere be part we have to calculate total amount of energy release if all hydrogen burns let us start solving it mass of the sphere you can find volume into density density is given 0 .2 gram per centimeter cube volume will be 4 by 3 pi rq radius is given 1 .5 into 10 to the power minus 2.

03:27 On solving it it would be 3 .53 10 to the power minus 7 gram.

03:36 So this is the mass of hydrogen sphere.

03:41 Now number of atoms we are finding mass upon molecular mass into abogadro number.

04:11 Since it having equal number of 1h2 and 1s3 so we can write molecular mass of 1h2 plus molecular mass of 1h3 divided by 2 into avogadro number now substituting the value mass of the hydrogen sphere is 3 .53 10 to the power minus 7 abogadro number 6 .023 10.

04:45 To the power 23 molecular mass 2 plus 3 upon 2.

04:53 So on solving it in this is in gram number of atoms in the sphere we will get 8 .51 into 10 to the power 16 atoms we have to find the temperature temperature can be defined as in terms of energy release energy absorb is 1 % of 200 kilojoule that is 2 kilojoule so 2 ,000 joule and this will be skull to 3 by 2 k t into 2n so temperature you will get 2 ,000 divided by 3 k t substitute 3 k n k is boltzman constant 1 .38 10 to the power minus 23 k t 8 .51 10 to the power 16.

06:30 So temperature you will get 5 .68 into 10 to the power 8 kelvin b part.

06:48 Each fusion reaction uses two nuclei...

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