A positive, singly ionized atom of mass number $ A_M $ is accelerated from rest by the voltage $ 192 \, \text{V} $. Thereafter, it enters a rectangular region of width $ w $ with magnetic field $ \vec{B}_0 = 0.1\hat{k} \, \text{T} $. The ion finally hits a detector at the distance $ x $ below its starting trajectory. Which of the following option(s) is(are) correct?
$ \text{(Given: Mass of neutron/proton = } \frac{5}{3} \times 10^{-27} \, \text{kg, charge of the electron = } 1.6 \times 10^{-19} \, \text{C).} $
A positive, singly ionized atom of mass number $ A_M $ is accelerated from rest by the voltage $ 192 \, \text{V} $. Thereafter, it enters a rectangular region of width $ w $ with magnetic field $ \vec{B}_0 = 0.1\hat{k} \, \text{T} $. The ion finally hits a detector at the distance $ x $ below its starting trajectory. Which of the following option(s) is(are) correct?
$ \text{(Given: Mass of neutron/proton = } \frac{5}{3} \times 10^{-27} \, \text{kg, charge of the electron = } 1.6 \times 10^{-19} \, \text{C).} $
Show Hint
- The value of \( x \) for \( \text{H}^+ \) ion is \( 4 \, \text{cm}. \)
- The value of \( x \) for an ion with \( A_M = 144 \) is \( 48 \, \text{cm}. \)
- For detecting ions with \( 1 \leq A_M \leq 196 \), the minimum height \( (x_1 - x_0) \) of the detector is \( 55 \, \text{cm}. \)
- The minimum width \( w \) of the region of the magnetic field for detecting ions with \( A_M = 196 \) is \( 56 \, \text{cm}. \)
The Correct Option is A
Approach Solution - 1
The displacement: \[ x = 2R = \frac{2mv}{qB}. \] Substituting: \[ x = \frac{2\sqrt{2m(qV)}}{qB}. \] For \( \text{H}^+ \), \( x = 4 \, \text{cm} \). For \( A_M = 144 \): \[ x \propto \sqrt{m} \quad \Rightarrow \quad x = 48 \, \text{cm}. \]
Approach Solution -2
The correct options are (A) and (B).
Step 1: The formula is given by:
\[ x = 2R = \frac{2mv}{qB} = 2 \sqrt{\frac{2m(e \Delta V)}{qB}} \]
For H+ ion, we have:
\[ x = 3.91 \, \text{cm} \quad \text{and} \quad x = 4 \, \text{cm}. \] Since (A) is correct, we move on to the next calculation.
Step 2: For m = 144 (mass of the proton), we get:
\[ x = 12 (x_{H^+}) = 48 \, \text{cm}. \] This shows that option (B) is correct.
Step 3: For the condition where 1 ā¤ AM ā¤ 196, we use the formula:
\[ (x_1 - x_0)_{\text{min}} = 2R_{196} - 2R_1 = (14 \times 4) - 4 = 52 \, \text{cm}. \] Thus, option (C) is incorrect.
Step 4: For AM = 196, the minimum width is given by:
\[ w_{\text{min}} = R_{196} = 28 \, \text{cm}. \] This shows that option (D) is incorrect.
Thus, the correct options are (A) and (B).
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