The mass of $_{7} N^{15}$ is $15.0011 a m u$ , mass of $_{8} O^{16}$ is $15.99492 a m u$ and $m_{P} = 1.00783 a m u$ . Determine binding energy of last proton of $_{8} O^{16}$ .

2.13 M e V

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0.13 M e V

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10 M e V

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12.13 M e V

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Solution

The correct option is D $12.13 M e V$
$M (_{8} O_{16}) = M (_{7} N^{15}) + 1 m_{P}$
binding energy of last proton
$= M (N^{15}) + m_{P} - M (_{1} O^{16})$
$= 15.00011 + 1.00783 - 15.99492$
$= 0.01302 a m u = 12.13 M e V$ .

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