PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 29, Problem 82P

(a)

To determine

The closest approach an α particle with a kinetic energy 1.0 MeV to the gold nucleus.

(a)

Expert Solution
Check Mark

Answer to Problem 82P

The closest approach an α particle with a kinetic energy 1.0 MeV to the gold nucleus is 230 fm.

Explanation of Solution

An α particle is approaching the gold nucleus with a kinetic energy Kα=1.0 MeV and the charge of α particle is qα=2e and gold nucleus is qAu=79e. The charge e=1.602×1019 C and the kinetic energy of α particle, Kα=UE.

Write the formula for electric potential energy

    UE=kqαqAur                                                                                     (I)

Here, UE is the electric potential energy, k is Coulomb constant, qα&qAu is the charge and r is the radius.

Rewrtie equation (I) to solve for r

r=kqαqAuKα

Substitute 1.0 MeV for Kα, 8.988×109 Nm2/C2 for k, 2e for qα and 79e for qAu in above equation to find the value of r

r=8.988×109 Nm2/C2×2e×79e1.0 MeVr=8.988×109 Nm2/C2×158×1.602×1019 C1.0×106 eV×1.602×1019 Cr=230 fm

Thus, the closest approach an α particle with a kinetic energy 1.0 MeV to the gold nucleus is 230 fm.

(b)

To determine

Will an α particle with a kinetic energy 1.0 MeV get close enough to touch the gold nucleus.

(b)

Expert Solution
Check Mark

Answer to Problem 82P

An α particle with a kinetic energy 1.0 MeV will not get close enough to touch the gold nucleus.

Explanation of Solution

The closest approach an α particle with a kinetic energy 1.0 MeV to the gold nucleus is 230 fm. The mass number of gold is AAu=197 and the α particle is Aα=4.

Write the equation for radius

    r=r0A1/3                                                                              (I)

Here, r is the radius, r0 is the initial radius r0=1.2 fm and A is the mass number.

The radius of α particle is r=1.2 fm×41/3=1.9 fm

The radius of gold nucleus is r=1.2 fm×1971/3=7.0 fm

Therefore, the distance between the centers of α particle and gold nucleus when they just touching is 8.90 fm. And the closest approach an α particle with a kinetic energy 1.0 MeV to the gold nucleus is 230 fm. Thus, an α particle with a kinetic energy 1.0 MeV will not get close enough to touch the gold nucleus

(c)

To determine

The minimum value of the total kinetic energy of  α particle and gold nucleus necessary to bring them into contact.

(c)

Expert Solution
Check Mark

Answer to Problem 82P

The minimum value of the total kinetic energy of  α particle and gold nucleus necessary to bring them into contact is 26 MeV.

Explanation of Solution

The distance between the centers of proton and nitrogen nucleus when they just touching is 8.90 fm and the charge of α particle is qα=2e and gold nucleus is qAu=79e. The charge e=1.602×1019 C

Substitute 8.90 fm for r, 8.988×109 Nm2/C2 for k, 2e for qα and 79e for qAu in equation (I) to find the value of UE

UE=8.988×109 Nm2/C2×2e×79e8.90 fmUE=8.988×109 Nm2/C2×158×1.602×1019 C8.90×1015 m×1 eV1.602×1019 JUE=26 MeV

Therefore, the minimum value of the total kinetic energy of  α particle and gold nucleus necessary to bring them into contact is 26 MeV.

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Chapter 29 Solutions

PHYSICS

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