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Chapter 18: Problem 2472

It the radius of a nucleus of mass number 3 is \(\mathrm{R}\). then the radius of a nucleus of mass number 81 is (A) \(27 \mathrm{R}\) (B) \(9 \mathrm{R}\) (C) \(3 \mathrm{R}\) \((\mathrm{D})(27)^{(1 / 2)} \mathrm{R}\)

Short Answer

Expert verified
The radius of a nucleus with mass number 81 is (C) \(3 \mathrm{R}\).

Step by step solution

01

Write down the formula for the radius of the nucleus with mass number 3.

We know the radius of a nucleus with mass number 3 is R. Using the formula, we have: \( R = R_0 (3)^{\frac{1}{3}} \)
02

Write down the formula for the radius of the nucleus with mass number 81.

We want to find the radius of a nucleus with mass number 81. Using the formula, we have: \( R_n = R_0 (81)^{\frac{1}{3}} \)
03

Divide the equation for mass number 81 by the equation for mass number 3.

By dividing both equations, we can eliminate the constant \(R_0\), allowing us to find the relationship between the radii of the nuclei with mass numbers 3 and 81: \(\frac{R_n}{R} = \frac{R_0 (81)^{\frac{1}{3}}}{R_0 (3)^{\frac{1}{3}}} \)
04

Simplify the equation.

The constant \(R_0\) can now be cancelled out, and we can simplify the expression as follows: \(\frac{R_n}{R} = \frac{(81)^{\frac{1}{3}}}{(3)^{\frac{1}{3}}} = (27)^{\frac{1}{3}} = 3\)
05

Find the desired radius.

Finally, we multiply both sides of the equation by the given radius R: \(R_n = 3R\) Comparing our result with the given options, we find that the correct answer is (C) \(3 \mathrm{R}\).

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Most popular questions from this chapter

If a hydrogen atom emits a Photon of wave length \(\lambda\). the recoil speed of the atom of mass \(\mathrm{m}\) is given by (A) \((\mathrm{h} / \mathrm{m} \lambda)\) (B) \((\mathrm{mh} / \lambda)\) (C) \(\operatorname{mh} \lambda\) (D) \((\mathrm{m} \lambda / \mathrm{h})\)

If \(13.6 \mathrm{eV}\) energy is required to ionise the hydrogen atom the energy required to remove the electron form \(n=2\) state is (A) Zero (B) \(10.2 \mathrm{eV}\) (C) \(6.8 \mathrm{eV}\) (D) \(3.4 \mathrm{eV}\)

when \({ }_{3}^{7}\) Li nuclear are bombarded by Proton and the resultant nuclei are \({ }^{8}{ }_{4}\) Be, the emitted particle will be (A) neutron (B) gamma (C) alpha (D) Beta

The energy difference between the first two levels of hydrogen atom is $10.2 \mathrm{eV}$. what is the corresponding energy difference for a singly ionized helium atom? (A) \(10.2 \mathrm{eV}\) (B) \(81.6 \mathrm{eV}\) (C) \(20.4 \mathrm{eV}\) (D) \(40.8 \mathrm{eV}\)

The binding energy Per nucleon of deuteron $\left({ }^{2}{ }_{1} \mathrm{H}\right)\( and Lielium nucleus \){ }_{2}{ }^{4}{ }_{2} \mathrm{He}$ ) is \(1.1 \mathrm{MeV}\) and \(7.0 \mathrm{MeV}\). respectively. If two deuteron react to form a single helium nucleus, the energy released is (A) \(23.6 \mathrm{MeV}\) (B) \(26.9 \mathrm{MeV}\) (C) \(13.9 \mathrm{MeV}\) (D) \(19.2 \mathrm{MeV}\)
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