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
Answers
Explanation:
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⟹% Change in mass no: of
6
C
12
=
Original Mass number
Change in Mass number
×100
\boxed{\sf Change \ in \ mass \ number = Original \ mass \ number - New \ mass \ number}
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⟹% Change in mass no: of
6
C
12
=
12
12−9
×100
\sf \Longrightarrow \% \ Change \ in \ mass \ no: \ of \ _{6}C^{12} \ = \dfrac{3}{12} \times 100⟹% Change in mass no: of
6
C
12
=
12
3
×100
\sf \Longrightarrow \% \ Change \ in \ mass \ no: \ of \ _{6}C^{12} \ = \dfrac{1}{4} \times 100⟹% Change in mass no: of
6
C
12
=
4
1
×100
\sf \Longrightarrow \% \ Change \ in \ mass \ no: \ of \ _{6}C^{12} \ = 25 \%⟹% Change in mass no: of
6
C
12
=25%
Final answer: 25%
Answer:
in
Explanation:
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