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Chapter 18: Problem 2487

It the radius of \({ }^{27}{ }_{13} \mathrm{~A} \ell\) nucleus is $3.6 \mathrm{fm}\( the radius of \){ }^{125}{ }_{52} \mathrm{Te}$ nucleus is nearly equal to (A) \(8 \mathrm{fm}\) (B) \(6 \mathrm{fm}\) (C) \(4 \mathrm{fm}\) (D) \(5 \mathrm{fm}\)

Short Answer

Expert verified
The radius of the \({ }^{125}{ }_{52} \mathrm{Te}\) nucleus is approximately \(6 \mathrm{fm}\).

Step by step solution

01

Understand the formula for Nuclear Radius

The formula for nuclear radius is given by: \[R = R_0 A^{1/3}\] where R is the nuclear radius, R_0 is the constant (approximately equal to 1.2 \(\mathrm{fm}\)), and A is the mass number of the nucleus.
02

Identify the given values

The problem gives us the radius of \({ }^{27}{ }_{13} \mathrm{~A} \ell\) nucleus as \(3.6 \mathrm{fm}\). So, in this case, A = 27 and R = 3.6 \(\mathrm{fm}\).
03

Calculate the value of R_0

To find the value of R_0, we need to rearrange the nuclear radius formula and substitute the values of A and R: \[R_0 = \frac{R}{A^{1/3}}\] Plug in the given values: \[R_0 = \frac{3.6}{27^{1/3}}\] \[R_0 \approx 1.2 \mathrm{fm}\]
04

Calculate the radius of \({ }^{125}{ }_{52} \mathrm{Te}\) nucleus

To find the radius of \({ }^{125}{ }_{52} \mathrm{Te}\) nucleus, use the nuclear radius formula. Assign the values A = 125, and R_0 = 1.2 \(\mathrm{fm}\). \[R = R_0 A^{1/3}\] \[R = 1.2 (125)^{1/3}\] \[R \approx 6 \mathrm{fm}\]
05

Identify the correct option

After calculating the radius of \({ }^{125}{ }_{52} \mathrm{Te}\) nucleus, compare it with the given options. Since the radius is approximately \(6 \mathrm{fm}\), the correct answer is (B) \(6 \mathrm{fm}\).

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Most popular questions from this chapter

Match column I and II and chose correct Answer form the given below. (a) Nuclear fusion (p) converts some matter into energy (b) Nuclear fission (q) generally Possible for nuclei with low atomic number (c) \(\beta\) decay (r) generally Possible for nuclei with high atomic number (d) Exothermic nuclear (s) Essentially Proceeds by weak reaction nuclear force(c) (A) $\mathrm{a} \rightarrow \mathrm{p}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{q}$ (B) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{d} \rightarrow \mathrm{s}$ (C) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{p}$ (D) $\mathrm{a} \rightarrow \mathrm{r}, \mathrm{b} \rightarrow \mathrm{q}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{d} \rightarrow \mathrm{s}$

which of the following statement is true (A) \({ }_{78} \mathrm{Pt}^{192}\) has 78 neutrons (B) ${ }_{90} \mathrm{Th}^{234} \rightarrow{ }_{91} \mathrm{~Pa}^{234}+{ }_{2} \mathrm{He}^{4}$ (C) ${ }_{92} \mathrm{U}^{238} \rightarrow{ }_{90} \mathrm{Th}^{234}+{ }_{2} \mathrm{He}^{4}$ (D) ${ }_{84} \mathrm{Po}^{214} \rightarrow{ }_{82} \mathrm{~Pb}^{210}+\beta^{-}$

what Percent of original radioactive substance is left after 5 half life time (A) \(3 \%\) (B) \(5 \%\) (C) \(6 \%\) (D) \(12 \%\)

In gamma ray emission form a nucleus (A) there is no change in the proton-number and neutron number (B) Both the number are changes (C) only Proton number change (D) only neutron number change

which of these is a fusion reaction (A) ${ }^{3}{ }_{1} \mathrm{H}+{ }_{1}^{2} \mathrm{H}={ }_{2}^{4} \mathrm{He}+{ }_{0} \mathrm{n}^{1}$ (B) ${ }^{12}{ }_{7} \mathrm{C} \rightarrow{ }_{6} \mathrm{C}^{12}+\beta^{+}+\mathrm{g}$ (C) ${ }_{92} \mathrm{U}^{238} \rightarrow{ }^{206}{ }_{82} \mathrm{~Pb}+8\left({ }_{2}^{4} \mathrm{He}\right)+6\left(_{1-} \mathrm{e}\right)$ (D) None of these
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