Answer:
Explanation:
(a) For this problem we can use the expression
[tex]v=\frac{qBr}{m}[/tex]
where B is the magnetic field, r is the radius of the curve described by the ion, m is the mass and q is the electric charge.
In this case, both isotopes have the same velocity, and we can assume that both isotopes has been ionized to have the same charge. Being r and r' the radius of the curve for 12C ion and 14C ion respectively, we have
[tex]v=v'\\\frac{qBr}{m}=\frac{qBr'}{m'}\\r'=\frac{m'r}{m}=\frac{(14u)(21.15cm)}{(12u)}=24.67cm\\[/tex]
Hence, 2*24.67cm=49.35cm is the diameter of the orbit for 14C.
(b) we can calculate Ļ by using
[tex]\omega_{12C} =\frac{v}{r}\\\omega_{14C}=\frac{v}{r'}\\[/tex]
By dividing these expressions we have
[tex]\frac{\omega_{12C}}{\omega_{14C}}=\frac{r'}{r}=\frac{24.67cm}{21.15cm}=1.16[/tex]
The isotope 12C has 1.16 more angular frecuency than 14C
I hope this is useful for you
regards
