The mass of _ 7 N^ 15 is 15.00011 amu, mass of _ 8 O^ 16 is 15.99492 amu and m_ p =1.00783 amu. Determine binding energy of last proton of _ 8 O^ 16 .

Correct answer: D. 12.13 MeV 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 amu=12.13 MeV

BITSAT 2016

Exam: BITSAT
Year: 2016
Subject: Physics
Chapter: Nuclear Physics

BITSAT 2016 · Physics · Nuclear Physics

The mass of _ 7 N^ 15 is 15.00011 amu, mass of _ 8 O^ 16 is...

The mass of _{7}

\mathrm{ N}^{15} is 15.00011 \mathrm{amu}, mass of _{8} \mathrm{O}^{16}

is 15.99492 amu and

m_{p}=1.00783

amu. Determine binding energy of last proton of

{ }_{8} \mathrm{O}^{16}

Options

A $2.13 \mathrm{MeV}$
B $0.13 \mathrm{MeV}$
C $10 \mathrm{MeV}$
D $12.13 \mathrm{MeV}$

Correct answer

D. $12.13 \mathrm{MeV}$

Step-by-step solution

\mathrm{M}\left({ }_{8} \mathrm{O}^{16}\right)=\mathrm{M}\left({ }_{7} \mathrm{ N}^{15}\right)+1 \mathrm{ m}_{\mathrm{p}}

binding energy of last proton

=M\left(N^{15}\right)+m_{P}-M\left({ }_{1} O^{16}\right)=15.00011+1.00783-15.99492=0.01302 \mathrm{amu}=12.13 \mathrm{MeV}

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