A narrow beam of singly-charged carbon ions travelling at a...

When a charged particle enters a uniform magnetic field perpendicularly, it moves in a circular path of radius $r = \dfrac{mv}{qB}$.

The separation between the incident beam and the emergent beam in the backward direction is the diameter of the circular path, $d = 2r = \dfrac{2mv}{qB}$.

Substituting the given values:
$d = \dfrac{2 \times (A \times 1.6 \times 10^{-27}) \times 6.0 \times 10^4}{1.6 \times 10^{-19} \times 0.5}$

$d = 24 \times 10^{-4} A \text{ m} = 0.24 A \text{ cm}$

For the first beam, $d_1 = 3.0 \text{ cm}$:
$A_1 = \dfrac{3.0}{0.24} = 12.5$

For the second beam, $d_2 = 3.5 \text{ cm}$:
$A_2 = \dfrac{3.5}{0.24} \approx 14.6$

Since the mass number $A$ must be an integer, the closest valid carbon isotopes are $12$ and $14$. Using the standard atomic mass unit $1 \text{ amu} \approx 1.66 \times 10^{-27} \text{ kg}$ instead of the given approximation yields $d \approx 0.25 A \text{ cm}$, which gives exactly $A_1 = 12$ and $A_2 = 14$.

Thus, the isotopes present are $^{12}\text{C}$ and $^{14}\text{C}$.

Answer: $^{12}\text{C}$ and $^{14}\text{C}$