The binding energy per nucleon of 4020Ca is 4020Ca -39-962589 amu- mp -1-007825 amu- mn- 1-008665 amu and 1amu is equivalent to 931-5 MeV

The correct option is B 8.55 MeV Δ m = [20× 1.007825 + 20× 1.008665 39.962589] μ #160; #160; #160; #160; = 0.367211 μ BE = 0. ...

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Standard XII

Physics

Power, Force and Velocity

The binding e...

Question

The binding energy per nucleon of $_{20}^{40} C a$ is $- ------- -$
( $_{20}^{40} C a =$ 39.962589 amu, $m_{p} =$ 1.007825 amu; $m_{n} =$ $1.008665 a m u$ and $1 a m u$ is equivalent to $931.5 M e V$ )

6.55 M e V

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7.55 M e V

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8.55 M e V

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9.55 M e V

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Solution

The correct option is B $8.55 M e V$
$Δ m = [20 \times 1.007825 + 20 \times 1.008665 - 39.962589] μ$
$= 0.367211 μ$
$B E = 0.367211 \times 931.478 M e V$
$= 342.045 M e V$
BE per nucleon $= \frac{B E}{40} = 8.55 M e V$

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Similar questions

Q. The binding energy per nucleon of

_{17}^{35} C l

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(

_{17}^{35} C l =

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m_{P} =

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m_{n} =

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Q. The mass defect and binding energy of

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(Mass of

_{6}^{12}

=

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m_{p} =

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Q. Calculate the nuclear binding energy per nucleon of

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from the following data:
Given mass of proton

(m_{p}) = 1.00758

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mass of neutron

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mass of electron

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Isotopic mass of

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1 a m u = 931

MeV
Plan from the mass defect

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binding energy per nucleon

= \frac{△ m \times 931}{n u m b e r o f n u c l e o n s}

Q. Calculate the binding energy per nucleon for

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The binding energy per nucleon of 4020Ca is –––––––––(4020Ca=39.962589 amu, mp=1.007825 amu; mn=1.008665amu and 1amu is equivalent to 931.5MeV)

The correct option is B 8.55MeVΔm=[20×1.007825+20×1.008665−39.962589]μ =0.367211μBE=0.367211×931.478MeV =342.045MeVBE per nucleon =BE40=8.55 MeV

The binding energy per nucleon of $_{20}^{40} C a$ is $- ------- -$
( $_{20}^{40} C a =$ 39.962589 amu, $m_{p} =$ 1.007825 amu; $m_{n} =$ $1.008665 a m u$ and $1 a m u$ is equivalent to $931.5 M e V$ )

The correct option is B $8.55 M e V$
$Δ m = [20 \times 1.007825 + 20 \times 1.008665 - 39.962589] μ$
$= 0.367211 μ$
$B E = 0.367211 \times 931.478 M e V$
$= 342.045 M e V$
BE per nucleon $= \frac{B E}{40} = 8.55 M e V$