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Chapter 18: Problem 2499

In the following nuclear fusion reaction ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0} \mathrm{n}^{1}$ the repulsive potential energy between the two fusing nuclei is $7.7 \times 10^{-14} \mathrm{~J}$. The Temperature to which the gas must be heated is nearly (Boltzman constant \(\mathrm{K}=1.38 \times 10^{-23} \mathrm{JK}^{-1}\) ) (A) \(10^{3} \mathrm{~K}\) (B) \(10^{5} \mathrm{~K}\) (C) \(10^{7} \mathrm{~K}\) (D) \(10^{9} \mathrm{~K}\)

Short Answer

Expert verified
The temperature required to overcome the repulsive potential energy for this nuclear fusion reaction is around \(5.58 \times 10^{9} K\). Therefore, the closest answer is option (D) - \(10^{9} K\).

Step by step solution

01

Understand the problem

The problem is asking us to determine the minimal temperature required to overcome the repulsive potential energy for a nuclear fusion reaction. This is done by rearranging the formula of the most probable energy in a Maxwell-Boltzmann distribution, \(E = kT\) for some constant \(k\), to solve for \(T\) (Temperature).
02

Rearrange the formula to solve for T

We rearrange the formula \(E = kT\) to solve for \(T\). This gives us \(T = \frac{E}{k}\).
03

Substitute the values into the formula

We substitute the repulsive potential energy \(E = 7.7 \times 10^{-14} J\) and the Boltzmann constant \(k = 1.38 \times 10^{-23} \: JK^{-1}\) into the formula from step 2. When we do this, we get \(T = \frac{7.7 \times 10^{-14} \: J}{1.38 \times 10^{-23} \: JK^{-1}}\).
04

Calculate T (Temperature)

Evaluate \(T = \frac{7.7 \times 10^{-14} \: J}{1.38 \times 10^{-23} \: JK^{-1}}\) to find the temperature. A careful calculation yields \(T = 5.58 \times 10^{9} K\).
05

Make a conclusion

From our calculations, the temperature required to overcome the repulsive potential energy for this nuclear fusion reaction is around \(5.58 \times 10^{9} K\). This temperature is greater than all the options given except (D) - \(10^{9} K\). Hence, option D is the closest and is accepted as the answer.

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Most popular questions from this chapter

Match column I and II column I \(\frac{\text { column I }}{\text { fnucleus }}\) (a) size of (b) number of Proton in a nucleus column II (p) Z (q) \(10^{-15} \mathrm{~m}\) (r) \((\mathrm{A}-\mathrm{Z})\) (s) \(10^{-10} \mathrm{~m}\) 0 (c) size of Atom (d) Number of neutrons in a nucleus (A) $\mathrm{a} \rightarrow \mathrm{r}, \mathrm{b} \rightarrow \mathrm{s}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{d} \rightarrow \mathrm{p}$ (B) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{p}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{r}$ (C) $\mathrm{a} \rightarrow \mathrm{s}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{d} \rightarrow \mathrm{p}$ (D) $\mathrm{b} \rightarrow \mathrm{s}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{d} \rightarrow \mathrm{r}$

Nucleon is common name for (A) electron and neutron (B) proton and neutron (C) neutron and positron (D) neutron and neutrino

\(\mathrm{A}\) and \(\mathrm{B}\) are two radioactive substance whose half lives are 1 and 2 years respectively. Initially \(10 \mathrm{~g}\) of \(\mathrm{A}\) and \(1 \mathrm{~g}\) of \(\mathrm{B}\) is taken. The time after which they will have same quantity remaining is (A) \(3.6\) years (B) 7 years (C) \(6.6\) years (D) 5 years

The Probability of survival of a radioactive nucleus for one mean life time is (A) \(1-(1 / \mathrm{e}) \mathrm{S}\) (B) \((1 / \mathrm{e})\) (C) \((2 / \mathrm{e})\) (D) \((3 / \mathrm{e})\)

In terms of Rydberg constant \(R\). The wave number of first Balmer line is (A) \((5 \mathrm{R} / 36)\) (B) \((8 \mathrm{R} / 9)\) (C) \(\mathrm{R}\) (D) \((8 \mathrm{R} / 20)\)
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