Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 44, Problem 3P

(a)

To determine

The distance of the closest approach of alpha particle and a gold nucleus.

(a)

Expert Solution
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Answer to Problem 3P

The distance of the closest approach of alpha particle and a gold nucleus is 455 fm.

Explanation of Solution

On a head-on collision between an alpha particle and a gold nucleus 79197Au. The initial energy of the alpha particle is 0.55 MeV and during the collision the gold nucleus remains at rest. Then at the distance of the closest approach, the kinetic energy of alpha particle must be equal to the electrostatic potential energy, Ki=Uf.

Write to formula for electrostatic potential energy

    U=keq1q2d                                                                         (I)

Write to formula for kinetic energy

    K=12mv2                                                                         (II)

Here, U is the potential energy, K is the kinetic energy, ke is the electrostatic constant [ke=8.99×109 Nm2/C2], q1 and q2 are the charge of the nucleus, d is the distance of separation, m is the mass of the particle and v is the velocity of the particle.

Conclusion:

Here, Ki=Uf

Ki=Uf=keq1q2dd=keq1q2Ki

Substituting 8.99×109 Nm2/C2 for ke, 1.60×1019 C for e and 0.500 MeV for Ki in the above equation to find the value of d. [Note: q1=2e, q2=79e and 1 MeV=1.60×1013 J]

d=(8.99×109 Nm2/C2)(2)(79)(1.60×1019 C)2(0.500 MeV)(1.60×1013 J/MeV)=4.55×1013 m=455×1015 m=455 fm

Thus, the distance of the closest approach of alpha particle and a gold nucleus is 455 fm.

(b)

To determine

The initial minimum initial speed for an alpha particle to approach as closed as 300 fm to the gold nucleus.

(b)

Expert Solution
Check Mark

Answer to Problem 3P

The initial minimum initial speed for an alpha particle to approach as closed as 300 fm to the gold nucleus is 6.05×106 m/s.

Explanation of Solution

Consider the closed distance between the alpha particle and the gold nucleus is 300 fm and the kinetic energy of alpha particle must be equal to the electrostatic potential energy, Ki=Uf. The mass of the alpha particle is 6.645×1027 kg.

Conclusion:

Equating (I) and (II),

Ki=Uf12mαvi2=keq1q2dvi2=2keq1q2mαdvi=2keq1q2mαd

Substituting 8.99×109 Nm2/C2 for ke, 1.60×1019 C for e, 6.645×1027 kg for mα and 300 fm for d in the above equation to find the value of vi. [Note: q1=2e and q2=79e]

vi=2(8.99×109 Nm2/C2)(2)(79)(1.602×1019 C)2(6.645×1027 kg)(300×1015 m)=6.05×106 m/s

Thus, the initial minimum initial speed for an alpha particle to approach as closed as 300 fm to the gold nucleus is 6.05×106 m/s.

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Chapter 44 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

Ch. 44 - Prob. 8OQCh. 44 - Prob. 9OQCh. 44 - Prob. 10OQCh. 44 - Prob. 11OQCh. 44 - Prob. 12OQCh. 44 - Prob. 13OQCh. 44 - Prob. 1CQCh. 44 - Prob. 2CQCh. 44 - Prob. 3CQCh. 44 - Prob. 4CQCh. 44 - Prob. 5CQCh. 44 - Prob. 6CQCh. 44 - Prob. 7CQCh. 44 - Prob. 8CQCh. 44 - Prob. 9CQCh. 44 - Prob. 10CQCh. 44 - Prob. 11CQCh. 44 - Prob. 12CQCh. 44 - Prob. 13CQCh. 44 - Prob. 14CQCh. 44 - Prob. 15CQCh. 44 - Prob. 16CQCh. 44 - Prob. 17CQCh. 44 - Prob. 1PCh. 44 - Prob. 2PCh. 44 - Prob. 3PCh. 44 - Prob. 4PCh. 44 - Prob. 5PCh. 44 - Prob. 6PCh. 44 - Prob. 7PCh. 44 - Prob. 8PCh. 44 - Prob. 9PCh. 44 - Prob. 10PCh. 44 - Prob. 11PCh. 44 - Prob. 12PCh. 44 - Prob. 13PCh. 44 - Prob. 14PCh. 44 - Prob. 15PCh. 44 - Prob. 16PCh. 44 - Prob. 17PCh. 44 - Prob. 18PCh. 44 - Prob. 19PCh. 44 - Prob. 20PCh. 44 - Prob. 21PCh. 44 - Prob. 22PCh. 44 - Prob. 23PCh. 44 - Prob. 24PCh. 44 - Prob. 25PCh. 44 - Prob. 26PCh. 44 - Prob. 27PCh. 44 - Prob. 28PCh. 44 - Prob. 29PCh. 44 - Prob. 31PCh. 44 - Prob. 32PCh. 44 - Prob. 33PCh. 44 - Prob. 34PCh. 44 - Prob. 35PCh. 44 - Prob. 36PCh. 44 - Prob. 37PCh. 44 - Prob. 38PCh. 44 - Prob. 39PCh. 44 - Prob. 40PCh. 44 - Prob. 41PCh. 44 - Prob. 42PCh. 44 - Prob. 43PCh. 44 - Prob. 44PCh. 44 - Prob. 45PCh. 44 - Prob. 46PCh. 44 - Prob. 47PCh. 44 - Prob. 48PCh. 44 - Prob. 49PCh. 44 - Prob. 50PCh. 44 - Prob. 51PCh. 44 - Prob. 52PCh. 44 - Prob. 53PCh. 44 - Prob. 54APCh. 44 - Prob. 55APCh. 44 - Prob. 56APCh. 44 - Prob. 57APCh. 44 - Prob. 58APCh. 44 - Prob. 59APCh. 44 - Prob. 60APCh. 44 - Prob. 61APCh. 44 - Prob. 62APCh. 44 - Prob. 63APCh. 44 - Prob. 64APCh. 44 - Prob. 65APCh. 44 - Prob. 66APCh. 44 - Prob. 67APCh. 44 - Prob. 68APCh. 44 - Prob. 69APCh. 44 - Prob. 70APCh. 44 - Prob. 71APCh. 44 - Prob. 72APCh. 44 - As part of his discovery of the neutron in 1932,...Ch. 44 - Prob. 74APCh. 44 - Prob. 75APCh. 44 - Prob. 76APCh. 44 - Prob. 77CPCh. 44 - Prob. 78CP
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