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University Physics Volume 3

17th Edition
William Moebs + 1 other
ISBN: 9781938168185
BuyFindarrow_forward

University Physics Volume 3

17th Edition
William Moebs + 1 other
ISBN: 9781938168185
Textbook Problem

(a) Calculate the energy released in the a decay of 238U . (b) What fraction of the mass of a single 238U is destroyed in the decay? The mass of 234Th is 234.043593 u. (c) Although the fractional mass loss is large for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?

To determine

(a)

Energy released during alpha decay of 238U.

Explanation

Given data:

Mass of 92238U146=238.0507u

Mass of 90234Th144=234.043u

Mass of alpha particle = 4.0026u

1u = 931.5MeV/c2

Formula used:

  • Mass defect is the difference between the total mass of the nucleus and mass of individual nuclear.
  • Mass of nucleus is always less than the mass of protons & neutrons in the nucleus.
  • Mass defect is given as,
  •   Δm=Zmp+(AZ)mnmnuc

Where

    Δm = mass defect

    Zmp = Total mass of proton.

    (AZ)mn =Total mass of neutron.

      mnuc = Mass of californium.

  • The energy released can be determined by
  •   E=Δmc2

    Where Δm = mass defect

    c = light velocity

To determine

(b)

Fraction of mass destroyed during alpha decay is 238U.

To determine

(c)

Difficulty in observing macroscopic sample of U-238.

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