Verified Solution • 2025–26 Session moderate

QPages 230-234: Q40: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q40 Solution: A narrow beam of singly-charged carbon ions, moving at a…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often stop at the raw ratio $A_1/A_2=6/7$ and incorrectly conclude the isotopes are $^6$C and $^7$C; they forget that carbon’s mass numbers must be near 12, so the ratio must be multiplied by the smallest integer that yields realistic integer mass numbers (here 2, giving $^{12}$C and $^{14}$C).

A narrow beam of singly-charged carbon ions, moving at a constant velocity of

6.0 \times 10^4 \text{ m s}^{-1}

, is sent perpendicularly in a rectangular region of uniform magnetic field

B = 0.5 \text{ T}

figure). It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being 3.0 cm and 3.5 cm. Identify the isotopes present in the ion beam. Take the mass of an ion =

A1.6 \times 10^{-27}) \text{ kg}

, where

A

is the mass number.

Step-by-Step Verified Solution

**Answer:** The isotopes are

^{12}\text{C}

and

^{14}\text{C}

. **Step-by-Step Explanation:** The separation from the incident beam is

2r

. So

2r_1 = 3.0 \text{ cm}

,

2r_2 = 3.5 \text{ cm}

.

r = \frac{m v}{q B}

. For singly charged ions,

m \propto r

.

m_1/m_2 = r_1/r_2 = 3.0/3.5 = 6/7

. With mass numbers,

A_1/A_2 = 6/7

. The isotopes are

^{12}\text{C}

and

^{14}\text{C}

.

All Questions in this Chapter

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