Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 26, Problem 91P

(a)

To determine

Speed of movement of Polonium-218 nucleus.

(a)

Expert Solution
Check Mark

Answer to Problem 91P

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 91P

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

Physics

Ch. 26 - Prob. 1CQCh. 26 - Prob. 2CQCh. 26 - Prob. 3CQCh. 26 - Prob. 4CQCh. 26 - Prob. 5CQCh. 26 - Prob. 6CQCh. 26 - Prob. 7CQCh. 26 - Prob. 8CQCh. 26 - Prob. 9CQCh. 26 - Prob. 10CQCh. 26 - Prob. 11CQCh. 26 - Prob. 12CQCh. 26 - Prob. 1MCQCh. 26 - Prob. 2MCQCh. 26 - Prob. 3MCQCh. 26 - Prob. 4MCQCh. 26 - 5. Which best describes the proper time interval...Ch. 26 - Prob. 6MCQCh. 26 - Prob. 7MCQCh. 26 - Prob. 8MCQCh. 26 - Prob. 9MCQCh. 26 - Prob. 1PCh. 26 - Prob. 2PCh. 26 - Prob. 3PCh. 26 - Prob. 4PCh. 26 - Prob. 5PCh. 26 - Prob. 6PCh. 26 - Prob. 7PCh. 26 - Prob. 8PCh. 26 - Prob. 9PCh. 26 - Prob. 10PCh. 26 - Prob. 11PCh. 26 - Prob. 12PCh. 26 - Prob. 13PCh. 26 - Prob. 14PCh. 26 - Prob. 15PCh. 26 - Prob. 16PCh. 26 - Prob. 17PCh. 26 - Prob. 18PCh. 26 - Prob. 19PCh. 26 - Prob. 20PCh. 26 - Prob. 21PCh. 26 - Prob. 22PCh. 26 - Prob. 23PCh. 26 - Prob. 24PCh. 26 - Prob. 25PCh. 26 - Prob. 26PCh. 26 - Prob. 27PCh. 26 - Prob. 28PCh. 26 - Prob. 29PCh. 26 - Prob. 30PCh. 26 - Prob. 31PCh. 26 - Prob. 32PCh. 26 - Prob. 33PCh. 26 - Prob. 34PCh. 26 - Prob. 35PCh. 26 - Prob. 36PCh. 26 - Prob. 37PCh. 26 - Prob. 38PCh. 26 - Prob. 39PCh. 26 - 40. A white dwarf is a star that has exhausted its...Ch. 26 - Prob. 41PCh. 26 - Prob. 42PCh. 26 - Prob. 43PCh. 26 - Prob. 44PCh. 26 - Prob. 45PCh. 26 - Prob. 46PCh. 26 - Prob. 47PCh. 26 - Prob. 48PCh. 26 - Prob. 49PCh. 26 - Prob. 50PCh. 26 - Prob. 51PCh. 26 - Prob. 52PCh. 26 - Prob. 53PCh. 26 - Prob. 54PCh. 26 - Prob. 55PCh. 26 - Prob. 56PCh. 26 - Prob. 57PCh. 26 - Prob. 58PCh. 26 - Prob. 59PCh. 26 - Prob. 60PCh. 26 - Prob. 61PCh. 26 - Prob. 62PCh. 26 - Prob. 63PCh. 26 - Prob. 64PCh. 26 - Prob. 65PCh. 26 - Prob. 66PCh. 26 - Prob. 67PCh. 26 - Prob. 68PCh. 26 - Prob. 69PCh. 26 - 70. At the 10.0 km long Stanford Linear...Ch. 26 - Prob. 71PCh. 26 - Prob. 72PCh. 26 - Prob. 73PCh. 26 - Prob. 74PCh. 26 - Prob. 75PCh. 26 - Prob. 76PCh. 26 - Prob. 77PCh. 26 - Prob. 78PCh. 26 - Prob. 79PCh. 26 - Prob. 80PCh. 26 - Prob. 81PCh. 26 - Prob. 82PCh. 26 - Prob. 83PCh. 26 - Prob. 84PCh. 26 - Prob. 85PCh. 26 - Prob. 87PCh. 26 - Prob. 86PCh. 26 - Prob. 89PCh. 26 - Prob. 88PCh. 26 - Prob. 90PCh. 26 - Prob. 92PCh. 26 - Prob. 91PCh. 26 - Prob. 94PCh. 26 - Prob. 93PCh. 26 - 96. The solar energy arriving at the outer edge of...Ch. 26 - Prob. 96PCh. 26 - Prob. 97PCh. 26 - Prob. 98P
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