/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 100 A narrow beam of singly charged ... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 100

A narrow beam of singly charged carbon ions, moving at a constant velocity of \(6 \times 10^{4} \mathrm{~m} / \mathrm{s}\) is sent perpendicularly in a rectangular region having uniform magnetic field \(B=0.5 \mathrm{~T}\). It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being \(3 \mathrm{~cm}\) and \(3.5 \mathrm{~cm}\). If mass of an ion \(=A\left(1.66 \times 10^{-27}\right) \mathrm{kg}\), where \(A\) is the mass number then isotopes present in beam are (A) \({ }^{11} \mathrm{C}\) (B) \({ }^{12} \mathrm{C}\) (C) \({ }^{13} \mathrm{C}\) (D) \({ }^{14} \mathrm{C}\)

Short Answer

Expert verified
The isotopes present in the beam are \({ }^{12} \mathrm{C}\) and \({ }^{14} \mathrm{C}\). The correct options are (B) and (D).

Step by step solution

01

Since the given separations are in centimeters, we need to convert them to meters to be consistent with the SI unit system. For the first separation: \( 3 cm = 3 \times 10^{-2} m\) For the second separation: \( 3.5 cm = 3.5 \times 10^{-2} m\) #Step 2: Calculate the mass numbers using the formula for radius#

We will now use the separations as radii and the given velocity and magnetic field strength to calculate the mass numbers \(A_1\) and \(A_2\). For the first separation, we get: \( r_1 = \dfrac{A_1(1.66 \times 10^{-27})v}{1.6 \times 10^{-19}B} \) Solving for \( A_1 \), we have: \( A_1 = \dfrac{r_1(1.6 \times 10^{-19})B}{(1.66 \times 10^{-27})v} \) For the second separation, we get: \( r_2 = \dfrac{A_2(1.66 \times 10^{-27})v}{1.6 \times 10^{-19}B} \) Solving for \( A_2 \), we have: \( A_2 = \dfrac{r_2(1.6 \times 10^{-19})B}{(1.66 \times 10^{-27})v} \) #Step 3: Substitute the given values and compute the mass numbers#
02

For the first separation (3 cm): \( A_1 = \dfrac{(3 \times 10^{-2})(1.6 \times 10^{-19})(0.5)}{(1.66 \times 10^{-27})(6 \times 10^{4})} \) \( A_1 \approx 12 \) For the second separation (3.5 cm): \( A_2 = \dfrac{(3.5 \times 10^{-2})(1.6 \times 10^{-19})(0.5)}{(1.66 \times 10^{-27})(6 \times 10^{4})} \) \( A_2 \approx 14 \) #Step 4: Identify the isotopes of carbon#

The mass numbers obtained correspond to two carbon isotopes: - Carbon-12: \(A_1 = 12\) - Carbon-14: \(A_2 = 14\) Therefore, the isotopes present in the beam are \({ }^{12} \mathrm{C}\) and \({ }^{14} \mathrm{C}\). The correct options are (B) and (D).

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Most popular questions from this chapter

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