If the binding energy of deuterium is 2-23 MeV- then the mass defect in amu is-A- 0-0024 MeVB- 0-0048 MeVC- 0-0036 MeVD- 0-0012 MeV

The correct option is A 0.0024 MeV Binding energy E = Δ m × 931 MeV Given, the binding energy of Deuterium = 2.23 MeV ⇒ Δ m = E/931 MeV = 2.23/931 =0.0024 amu ...

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Standard X

Physics

Nuclear Fusion

If the bindin...

Question

If the binding energy of deuterium is 2.23 MeV, then the mass defect (in amu) is?

0.0012 MeV

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0.0036 MeV

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0.0024 MeV

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0.0048 MeV

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Solution

The correct option is C
0.0024 MeV

Binding energy $E = Δ m \times 931 M e V$

Given, the binding energy of Deuterium = 2.23 MeV

$\Rightarrow Δ m = \frac{E}{931 M e V} = \frac{2.23}{931} = 0.0024 a m u$

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If the binding energy of deuterium is 2.23 MeV, then the mass defect (in amu) is?

The correct option is C 0.0024 MeV Binding energy E=Δm×931MeV Given, the binding energy of Deuterium = 2.23 MeV ⇒ Δm = E931 MeV =2.23931 =0.0024 amu

The correct option is C
0.0024 MeV

Binding energy $E = Δ m \times 931 M e V$

Given, the binding energy of Deuterium = 2.23 MeV

$\Rightarrow Δ m = \frac{E}{931 M e V} = \frac{2.23}{931} = 0.0024 a m u$