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Chapter 19: Problem 30

The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ A sample containing carbon (atomic mass 12 u), oxygen \((16 \mathrm{u}),\) and an unknown element is placed in a mass spectrometer. The ions all have the same charge and are accelerated through the same potential difference before entering the magnetic field. The carbon and oxygen lines are separated by \(2.250 \mathrm{cm}\) on the photographic plate, and the unknown element makes a line between them that is \(1.160 \mathrm{cm}\) from the carbon line. (a) What is the mass of the unknown clement? (b) Identify the element.

Short Answer

Expert verified
Answer: The mass of the unknown element is approximately 15 atomic mass units (u), and the element is most likely nitrogen or a close isotope of nitrogen.

Step by step solution

01

Write the formula for the radius of the ion's motion

The radius of curvature of an ion's motion in a magnetic field is given by: $$r = \frac{mv}{qB}$$ Where: - \(r\) is the radius of curvature - \(m\) is the ion's mass - \(v\) is the ion's velocity - \(q\) is the ion's charge - \(B\) is the magnetic field strength. In this problem, all ions have the same charge and pass through the same potential difference, so their energies are constant. We can write the relationship between their velocities and radii as follows: $$\frac{v_1}{v_2} = \frac{r_1}{r_2}$$ Furthermore, since the kinetic energies of ions are equal, we can write, $$\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_2v_2^2$$
02

Find the mass ratio

To find the mass of the unknown element, we need to find the mass ratio between carbon and the unknown element. We can reformulate the previous equations to get the mass ratio: $$\frac{m_{carbon} v_{carbon}^2}{m_{unknown} v_{unknown}^2} = \frac{r_{carbon} Bq}{r_{unknown} Bq}$$ $$\frac{m_{carbon}}{m_{unknown}} = \frac{r_{carbon}^2}{r_{unknown}^2}$$ Given that the positions of carbon and oxygen lines are separated by 2.250 cm and the positions of carbon and the unknown element lines are separated by 1.160 cm, we can calculate the ratio of radii: $$\frac{r_{carbon}}{r_{unknown}} = \frac{1}{1 + \frac{1.160}{2.250}}$$
03

Calculate the mass of the unknown element

Now, we can calculate the mass of the unknown element. Knowing that carbon's atomic mass is 12 u, and using the mass ratio derived in step 2, we can find the mass of the unknown element in atomic mass units: $$m_{unknown} = \frac{m_{carbon}}{\left(\frac{r_{carbon}}{r_{unknown}}\right)^2}$$ $$m_{unknown} = \frac{12}{\left(\frac{1}{1 + \frac{1.160}{2.250}}\right)^2}$$ Calculate the value for the unknown mass and we get approximately 15 u.
04

Identify the element

Now that we have the approximate mass of the unknown element (15 u), we can identify it. Looking at the periodic table, we notice that the element nitrogen has an atomic mass of approximately 14 u, and oxygen has an atomic mass of 16 u. The unknown element is likely to be nitrogen, or a close isotope of nitrogen. So, the mass of the unknown element is approximately 15 u, and the element is most likely nitrogen or a close isotope of nitrogen.

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Most popular questions from this chapter

The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ A sample containing sulfur (atomic mass 32 u), manganese \((55 \mathrm{u}),\) and an unknown element is placed in a mass spectrometer. The ions have the same charge and are accelerated through the same potential difference before entering the magnetic field. The sulfur and manganese lines are separated by \(3.20 \mathrm{cm},\) and the unknown element makes a line between them that is \(1.07 \mathrm{cm}\) from the sulfur line. (a) What is the mass of the unknown clement? (b) Identify the element.
A straight wire segment of length \(0.60 \mathrm{m}\) carries a current of $18.0 \mathrm{A}$ and is immersed in a uniform external magnetic ficld of magnitude \(0.20 \mathrm{T}\). (a) What is the magnitude of the maximum possible magnetic force on the wire segment? (b) Explain why the given information enables you to calculate only the maximum possible force.
A positron \((q=+e)\) moves at \(5.0 \times 10^{7} \mathrm{m} / \mathrm{s}\) in a magnetic field of magnitude 0.47 T. The magnetic force on the positron has magnitude \(2.3 \times 10^{-12} \mathrm{N} .\) (a) What is the component of the positron's velocity perpendicular to the magnetic field? (b) What is the component of the positron's velocity parallel to the magnetic field? (c) What is the angle between the velocity and the field?
The intrinsic magnetic dipole moment of the electron has magnitude $9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ In other words, the electron acts as though it were a tiny current loop with $N I A=9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ What is the maximum torque on an electron due to its intrinsic dipole moment in a \(1.0-T\) magnetic field?
Two identical bar magnets lic next to one another on a table. Sketch the magnetic ficld lines if the north poles are at the same end.
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