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Equivalent Resistance in Series Circuit

For the trans...

Question

For the transformation, $_{7} N^{14} + \dots ⟶_{6} C^{14} +_{1} H^{1}$

The bombarding particle is:

A

Proton

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B

Neutron

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C

Deutron

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D

Electron

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Solution

The correct option is A Neutron
$_{7} N^{14} +_{x} A^{y} ⟶_{6} C^{14} +_{1} H^{1}$

$7 + x = 6 + 1$
$x = 0$

$14 + y = 14 + 1$
$y = 1$

The element $_{x} A^{y}$ is neutron ( $_{0} n^{1}$ ).

Hence, option B is correct.

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