Explanation The α decay of 92 236 U is given as, 92 236 U → 90 232 Th + 2 4 He Write the expression to obtain the height of the coulomb barrier for α decay of 92 236 U which is given by the electrostatic potential energy. U α = 1 4 π ε 0 Q α Q Th ( r α + r Th ) (I) Here, U α is the electric potential energy of the α decay of 92 236 U , ε 0 is the permittivity in free space, Q α is the charge of nucleus of the α particle, Q Th is the change of the nucleus of thorium and r α is the radius of the nucleus of the α particle and r Th is the radius of the nucleus of the theorem. Write the expression to obtain the fission of 92 236 U which results into two palladium nuclei which is given by the electrostatic potential energy. U fission = 1 4 π ε 0 Q Pd Q Pd ( r Pd + r Pd ) (II) Here, U fission is the electric potential energy of the fission of 92 236 U , ε 0 is the permittivity in free space, Q Pd is the charge of nucleus of the palladium and r Pd is the radius of the nucleus of the palladium. Write the expression to obtain the radius of the nucleus. r = ( 1.2 × 10 − 15 m ) ( A ) 1 / 3 (III) Here, r is the radius of the nucleus and A is the atomic number. Conclusion: Substitute 4 for A in equation (III) to calculate r α . r α = ( 1.2 × 10 − 15 m ) ( 4 ) 1 / 3 Substitute 232 for A in equation (III) to calculate r Th . r α = ( 1.2 × 10 − 15 m ) ( 232 ) 1 / 3 Substitute 8.99 × 10 9 N ⋅ m 2 / C 2 for 1 / 4 π ε 0 , 2 ( 1.60 × 10 − 19 C ) for Q α , 90 ( 1.60 × 10 − 19 C ) for Q Th , ( 1.2 × 10 − 15 m ) ( 4 ) 1 / 3 for r α and ( 1.2 × 10 − 15 m ) ( 232 ) 1 / 3 for r Th in equation (I) to calculate U α . U α = ( 8.99 × 10 9 N ⋅ m 2 / C 2 ) [ 2 ( 1.60 × 10 − 19 C ) ] [ 90 ( 1.60 × 10 − 19 C ) ] ( 1.2 × 10 − 15 m ) ( 4 ) 1 / 3 + ( 1.2 × 10 − 15 m ) ( 232 ) 1 / 3 = ( 8.99 × 10 9 N ⋅ m 2 / C 2 ) ( 2 ) ( 90 ) ( 1.60 × 10 − 19 C ) 2 [ ( 4 ) 1 / 3 + ( 232 ) 1 / 3 ] ( 1.2 × 10 − 15 m ) (IV) Substitute 128 for A in equation (III) to calculate r Pd

Physics for Science and Engineering With Modern Physics, VI - Student Study Guide

4th Edition

ISBN: 9780132273244

Author: Doug Giancoli

Publisher: PEARSON

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Chapter 42, Problem 33P

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The ratio of the height of the Coulomb barrier for α decay to that for fission of 92236U.

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Chapter 42, Problem 32P

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Physics for Science and Engineering With Modern Physics, VI - Student Study Guide

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The ratio of the height of the Coulomb barrier for α decay to that for fission of 92 236 U .

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The ratio of the height of the Coulomb barrier for α decay to that for fission of 92236U.

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