For $_{6} C^{14}$ , mass of proton is $a$ , mass of neutron is $b$ and B.E is $X$ . Find the mass of a nucleus of $_{6} C^{14}$ ( $C$ = speed of light in vacuum)

6 a + 14 b - \frac{X}{C^{2}}

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6 a + 8 b + \frac{X}{C^{2}}

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6 a + 8 b - \frac{X}{C^{2}}

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6 a + 14 b + \frac{X}{C^{2}}

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Solution

The correct option is C $6 a + 8 b - \frac{X}{C^{2}}$
We know binding energy is given by
$B . E = Δ m . C^{2}$
where $Δ m$ is the mass defect
$B . E . = {(Mass of constituents) - (Mass of nucleus)} C^{2}$
$\frac{X}{C^{2}} = 6 a + 8 b - M_{N}$
$M_{N} = 6 a + 8 b - \frac{X}{C^{2}}$

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Similar questions

Use the identity $(x + a) (x + b) = x^{2} + (a + b) x + a b$ to find the following products:
$(a b c - 4) (a b c - 2)$

x = 6 a + 8 b + 9 c

y = 2 b - 3 a - 6 c

z = c - b + 3 a

x + y + z

6 a + 9 b + 4 c

x = 6 a + 8 b + 9 c

y = 2 b - 3 a - 6 c

z = c - b + 3 a

2 x - y - 3 z

6 a + 17 b + 21 c

(6 a - 8 b) - (- 4 a - 5 b)

Solve for $a$ : $- 3 (a + 4) + 5 (a - 2) = 6$

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