A Tritium (3^_1 H) nucleus with kinetic energy of 5.80 MeV is incident on a stat

Question

A Tritium (3^_1 H) nucleus with kinetic energy of 5.80 MeV is incident on a stationary target of phosphorus (30^_15 P); the nuclear reaction that results is written as 3^_1 H + 30^_15 P rightarrow 32^_16 S + 1^_0 n Some useful atomic masses are given as: Tritium 3^_1 H = 3.016049 u Phosphorus 30^_15 P = 29.978314 u Sulfur 32^_16 S= 31.972071 u Neutron 1^_0 H = 1.008665 u Hydrogen 1^_1 H = 1.007825 u find the radius of the nucleus 30^_15 P find the nuclear binding energy of the nucleus 30^_15 P (in MeV) If we ignore the recoil of the 32^_16 S nucleus, find the kinetic energy of the emitted neutron in MeV

Explanation / Answer

a )

the radius of the nucleus is R

R = R0 A(1/3)

R = 1.2 X 10-15 ( 30 )1/3

R = 3.72 X 10-15 m

b )

= ( 30 - 29.978 ) X 931.5

= 20.493 MeV

c )

= ( 3.0160 + 29.978 - 31.972 - 1.008 ) 931.5

= 0.013612 X 931.5

= 12.68 MeV

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