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NEET CHEMISTRYMedium

If the bombardment of α\alpha-particle on 714N^{14}_{7}N emits protons, then the new atom will be:

A

817O^{17}_{8}O

B

816O^{16}_{8}O

C

614C^{14}_{6}C

D

Ne

Step-by-Step Solution

An α\alpha-particle is a helium nucleus, represented as 24He^{4}_{2}He. A proton is represented as 11H^{1}_{1}H or 11p^{1}_{1}p. The nuclear transmutation reaction can be written as:

714N+24HeZAX+11H^{14}_{7}N + ^{4}_{2}He \rightarrow ^{A}_{Z}X + ^{1}_{1}H

According to the law of conservation of mass number and atomic number in a nuclear reaction: Conservation of mass number (superscripts): 14+4=A+1    A=1714 + 4 = A + 1 \implies A = 17 Conservation of atomic number (subscripts): 7+2=Z+1    Z=87 + 2 = Z + 1 \implies Z = 8

The element with atomic number Z=8Z = 8 is Oxygen (OO). Therefore, the newly formed atom ZAX^{A}_{Z}X is 817O^{17}_{8}O.

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