The mass defect and binding energy of 126C nucleus isMass of 126 C -12-000000 amu- mp-1-007825 amu and mn-1-008665 amu

The correct option is B 0.09894 amu; 92.1 MeVC = 6 neutrons + 6 protons Δ m = (6× 1.007825 + 6× 1.008665 12) #160; #160; #160; ...

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Standard XII

Physics

The Uncertainty Principle

The mass defe...

Question

The mass defect and binding energy of $_{6}^{12}$ C nucleus is
(Mass of $_{6}^{12}$ C $=$ 12.000000 amu; $m_{p} =$ 1.007825 amu and $m_{n} =$ 1.008665 amu)

0.06522

amu;

72.1

MeV

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0.09894

amu;

92.1

MeV

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0.05315

amu;

102.2

MeV

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0.05315

amu;

82.2

MeV

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Solution

The correct option is B $0.09894$ amu; $92.1$ MeV
C $=$ 6 neutrons + 6 protons
$Δ m = (6 \times 1.007825 + 6 \times 1.008665 - 12)$
$= 0.09894 a m u$
$E = (0.09894 \times 931.478 \frac{M e V}{C^{2}}) C^{2}$
$= 92.1 M e V$

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Similar questions

Q. Find Binding energy of an

α -

particle in

M e V

[m_{p r o t o n} = 1.007825 a m u, m_{n e u t r o n} = 1.008665 a m u, m_{h e l i u m} = 4.002800 a m u]

The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu, the binding energy of the helium nucleus will be [1 amu= 931 MeV]

Q. The binding energy per nucleon of

_{17}^{35} C l

nucleus is

(

_{17}^{35} C l =

34.98000 amu,

m_{P} =

1.007825 amu,

m_{n} =

1.008665 amu and 1 amu is equivalent to 931MeV)

Q. The binding energy of

_{17}^{35} C l

nucleus is 298 MeV. Find its atomic mass. Given, mass of a proton

(m_{p}) = 1.007825 a m u

, mass of neutron

(m_{n}) = 1.008665 a m u

Q. The binding energy per nucleon of

_{20}^{40} C a

- ------- -

(

_{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

)

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The mass defect and binding energy of 126C nucleus is(Mass of 126 C =12.000000 amu; mp=1.007825 amu and mn=1.008665 amu)

The correct option is B 0.09894 amu; 92.1 MeVC = 6 neutrons + 6 protonsΔm=(6×1.007825+6×1.008665−12) =0.09894 amuE=(0.09894×931.478MeVC2)C2 =92.1MeV

The mass defect and binding energy of $_{6}^{12}$ C nucleus is
(Mass of $_{6}^{12}$ C $=$ 12.000000 amu; $m_{p} =$ 1.007825 amu and $m_{n} =$ 1.008665 amu)

The correct option is B $0.09894$ amu; $92.1$ MeV
C $=$ 6 neutrons + 6 protons
$Δ m = (6 \times 1.007825 + 6 \times 1.008665 - 12)$
$= 0.09894 a m u$
$E = (0.09894 \times 931.478 \frac{M e V}{C^{2}}) C^{2}$
$= 92.1 M e V$