The reaction- 4Be9 - 2He4 - 6C12 - 0n1-shows the bombardment of Berilium nucleus with a-

The correct option is B both A and C α #160;particles #160; #160;consist of two protons and two neutrons bound together into a particle identical t ...

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Standard VII

Chemistry

Sub-Atomic Particles

The reaction;...

Question

The reaction;
$_{4} B e^{9} +_{2} H e^{4} \to_{6} C^{12} +_{0} n^{1}$ .
shows the bombardment of Berilium nucleus with a:

α

- particle

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β

- particle

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H e

nucleus

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both A and C

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Solution

The correct option is B both A and C
$α$ - particles consist of two protons and two neutrons bound together into a particle identical to a helium nucleus. Beryllium $_{4} B e^{9}$ releases a neutron under bombardment by alpha rays. Overall 2 protons and 1 neutron added to the daughter nuclei, the atomic number increases by 2 and atomic mass by 3 therefore the product nuclei form is $_{6} C^{12}$ .
$_{4} B e^{9} +_{2} H e^{4} \to_{6} C^{12} +_{0} n^{1}$

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The reaction;4Be9+2He4→6C12+0n1.shows the bombardment of Berilium nucleus with a:

The reaction;
$_{4} B e^{9} +_{2} H e^{4} \to_{6} C^{12} +_{0} n^{1}$ .
shows the bombardment of Berilium nucleus with a: