Question

In C12 atom if mass of e' is doubled and mass of proton is halved, then calculate the percentage
change in mass no. of C12
​

Answer 1

We have a C¹² atom. Let's note down the number of protons, neutrons and electrons down first.

Protons (p) ⇔ 6

Neutrons (n) ⇔ 6

Electrons (n) ⇔ 6

Mass number = Neutrons + Protons

Mass number = 6 + 6

Mass number = 12

Now, the mass of electrons is doubled, and the mass of protons are halved. Hence now the number of protons, neutrons & electrons will be:

Protons (p) ⇔ 3

Neutrons (n) ⇔ 6

Electrons (n) ⇔ 12

New mass number = Neutrons + Protons

New mass number = 6 + 3

New mass number = 9

Now, we are asked to find the % change in the mass number of C¹².

$\sf \Longrightarrow \% \ Change \ in \ mass \ no: \ of \ _{6}C^{12} \ = \dfrac{Change \ in \ Mass \ number}{Original \ Mass \ number} \times 100$

$\boxed{\sf Change \ in \ mass \ number = Original \ mass \ number - New \ mass \ number}$

$\sf \Longrightarrow \% \ Change \ in \ mass \ no: \ of \ _{6}C^{12} \ = \dfrac{12 - 9}{12} \times 100$

$\sf \Longrightarrow \% \ Change \ in \ mass \ no: \ of \ _{6}C^{12} \ = \dfrac{3}{12} \times 100$

$\sf \Longrightarrow \% \ Change \ in \ mass \ no: \ of \ _{6}C^{12} \ = \dfrac{1}{4} \times 100$

$\sf \Longrightarrow \% \ Change \ in \ mass \ no: \ of \ _{6}C^{12} \ = 25 \%$

Final answer: 25%