Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 26, Problem 93P

(a)

To determine

Speed of movement of Polonium-218 nucleus.

(a)

Expert Solution
Check Mark

Answer to Problem 93P

Speed of particle is 2.98×105m/s.

Explanation of Solution

Write the equation for law of conservation of energy for the given system.

  mRac2=mαc2+mPoc2+Kα+KPo                                                                             (I)

Here, mRa is the mass of radon-222 nucleus, c is the speed of light in vacuum, mα is the mass of alpha particle, mPo is the mass of polonium-218 nucleus, Kα is the kinetic energy of alpha particle, and KPo is the kinetic energy of polonium-218 nucleus.

Write the equation to find Kα.

  Kα=p22mα                                                                                                               (II)

Here, p is the momentum.

Write the equation to find KPo.

  KPo=p22mPo                                                                                                           (III)

Rewrite equation (I) by subtitling equations (II) and (III).

  mRac2=mαc2+mPoc2+p22mα+p22mPo(mRamαmPo)c2=p2(12mα+12mPo)

Rewrite the above expression in terms of p.

    p=(mRamαmPo)c2(12mα+12mPo)                                                                                     (IV)

Write the equation to find the speed of Polonium-218 nucleus.

    vPo=pmPo                                                                                                              (V)

Here, vPo is the speed of Polonium-218 nucleus.

Conclusion:

Substitute 221.97039u for mRa, 4.00151u for mα, and 217.96289u for mPo in equation (IV) to find p.

    p=(221.97039u4.00151u217.96289u)c2(12(4.00151u)+12(217.96289u))=(0.00599u)0.1272=(0.217u)c

Substitute (0.217u)c for p, 2.998×108m/s for c, and 217.96289 for mPo in equation (V) to find vPo.

  vPo=(0.217u)(2.998×108m/s)217.96289u=2.98×105m/s

Therefore, the speed of particle is 2.98×105m/s.

(b)

To determine

Speed of movement of alpha particle.

(b)

Expert Solution
Check Mark

Answer to Problem 93P

Speed of particle is 1.63×107m/s.

Explanation of Solution

Write the equation to find the speed of alpha particle.

    vα=pmα                                                                                                              (VI)

Here, vPo is the speed of alpha particle.

Conclusion:

Substitute (0.217u)c for p, 2.998×108m/s for c, and 4.00151u for mPo in equation (V) to find vPo.

  vPo=(0.217u)(2.998×108m/s)217.96289u=2.98×105m/s

Therefore, the speed of particle is 1.63×107m/s.

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Chapter 26 Solutions

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