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Chapter 18: Problem 2558

The radius of Ge nucleus is measured to be twice the radius of ${ }^{9}{ }_{4}$ Be. The number of nucleons in Ge are (A) 72 (B) 78 (C) 65 (D) 80

Short Answer

Expert verified
Using the nuclear radius formula, we find that the mass number of Germanium (Ge) is $A_{Ge} = 72$. Therefore, the number of nucleons in Ge is 72. The correct option is (A).

Step by step solution

01

Write down the formula for the nuclear radius

The formula for the nuclear radius (R) in terms of the mass number (A) is given by: \[R = R_0 A^{1/3}\] where \(R_0\) is a constant, approximately equal to 1.2 fm (fermi, a unit of length used in nuclear physics, 1 fermi = \(10^{-15}\)m).
02

Write down the given information

We are given that the radius of Ge nucleus (R_Ge) is twice the radius of \({ }^{9}_{4}\) Be nucleus (R_Be): \[R_{Ge} = 2 R_{Be}\]
03

Find the mass number of Beryllium

Beryllium is given as \({ }^{9}_{4}\) Be, so its mass number (number of protons and neutrons) is A_Be = 9.
04

Substitute the formula for the nuclear radius

Now, we substitute the formula for the nuclear radius for each element: \[R_0 A_{Ge}^{1/3} = 2 R_0 A_{Be}^{1/3}\]
05

Divide by R_0

Divide both sides by \(R_0\): \[A_{Ge}^{1/3} = 2 A_{Be}^{1/3}\]
06

Substitute the mass number of Beryllium

We can now substitute the mass number of Beryllium (A_Be = 9) in the equation: \[A_{Ge}^{1/3} = 2 (9)^{1/3}\]
07

Calculate the cube of the equation

Take the cube of both sides of the equation: \[(A_{Ge}^{1/3})^3 = [2 (9)^{1/3}]^3\]
08

Calculate the mass number of Germanium

Simplify the equation to find the mass number of Germanium: \[A_{Ge} = 2^3 \times 9 = 8 \times 9 = 72\]
09

Identify the correct option

The number of nucleons (protons and neutrons) in Ge is 72. The correct option is: (A) 72

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Most popular questions from this chapter

If \(\mathrm{M}_{0}\) is the mass of an isotope, ${ }^{17}{ }_{8} \mathrm{O}, \mathrm{M}_{\mathrm{p}}\( and \)\mathrm{M}_{\mathrm{n}}$ are the masses of a Proton and neutron respectively, the binding energy of the isotope is (A) \(\left(\mathrm{M}_{0}-8 \mathrm{M}_{\mathrm{p}}\right) \mathrm{C}^{2}\) (B) $\left(\mathrm{M}_{0}-8 \mathrm{M}_{p}-9 \mathrm{M}_{\mathrm{n}}\right) \mathrm{C}^{2}$ (C) \(\left(\mathrm{M}_{0}-17 \mathrm{M}_{\mathrm{n}}\right) \mathrm{C}^{2}\) (D) \(\mathrm{M}_{\mathrm{O}} \mathrm{C}^{2}\)

The ionization Potential of hydrogen atom is \(13.6 \mathrm{eV}\). An electron in the ground state absorbs Photon of energy \(12.75 \mathrm{eV}\). How many different spectral lines can one expect when electron make a down ward transition (A) 1 (B) 2 (C) 6 (D) 4

The half life time of a radioactive elements of \(\mathrm{x}\) is the same as the mean life of another radioactive element \(\mathrm{y}\). Initially they have same number of atoms, then (A) \(\mathrm{y}\) will decay faster then \(\mathrm{x}\) (B) \(\mathrm{x}\) will decay faster then \(\mathrm{y}\) (C) \(\mathrm{x}\) and \(\mathrm{y}\) will decay at the same rate at all time (D) \(\mathrm{x}\) and \(\mathrm{y}\) will decay at the same rate initially.

As the electron in Bohr is orbit of hydrogen atom Passes from state \(\mathrm{n}=2\) to \(\mathrm{n}=1\), the \(\mathrm{K} . \mathrm{E}\). and Potential energy changes as (A) Two fold, also two fold (B) four fold, two fold (C) four fold, also four fold (D) two fold, four fold

The binding energy Per nucleon of \({ }_{8} \mathrm{O}^{16}\) is $7.97 \mathrm{MeV}\( and that of \)_{8} \mathrm{O}^{17}\( is \)7.75 \mathrm{MeV}$ The energy \((\mathrm{in}-\mathrm{MeV})\) required to remove a neutron from $_{8} \mathrm{O}^{17}$ is (A) \(3.65\) (B) \(7.86\) (C) \(3.52\) (D) \(4.23\)
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