The binding energy per nucleon of Uranium in $M e V$ is
(Atomic mass of uranium $m_{a} = 238.0508 a m u;$
Mass of hydrogen atom is $M_{H} = 1.0078 a m u$ ;
Mass of neutron $m_{N} =$ $1.0087 a m u$ ; atomic number of Uranium $Z$ $=$ 92; Mass Number of Uranium $A$ $=$ 238)

7.580

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6.216

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5.162

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3.146

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Solution

The correct option is A $7.580$
No. of protons $=$ $92$

No. of neutrons $= 238 - 92 = 146$
$Δ m = [92 \times 1.0078 + 146 \times 1.0087 - 238.0508] a m u$
$= 1.937 a m u$
$B E = 1.937 \times 931.478 M e V$
$= 1804.27 M e V$
$B E / p e r n u c l e o n = \frac{1804.27}{238} M e V$
$= 7.58 M e V$

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α -

α

U^{238}

92

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β - p a r t i c l e

=

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