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$_{4} {Be}^{9}$ + $_{2} {He}^{4}$ $_{6} C^{12}$ + X find the X.

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_{0} n^{1}

_{0} n^{12}

_{0} n^{9}

_{0} n^{8}

answer is A.

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Detailed Solution

The value of X is given by

_{0} n^{1}

Let the particle be X.
On summing the mass number

LHS = 9 + 4

= 11 units

On RHS = 12 units

We should have unit mass number on x element.
On summing of atomic mass number on left hand side

LHS = 4 + 2

= 6 units

On right hand side atomic mass number o

RHS = 6 units

of atomic mass number
Number of .protons are balanced in with C (Atomic mass number)

LHS = RHS

6 = 6

Mass number is

LHS = RHS

11 = 12

11 < 12

11 - 12 = 1 unit

X carries mass number is 1 unit.
X carries no charges

_{4} {Be}^{9}

_{2} {He}^{4}

_{6} C^{12}

_{0} n^{1}

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