A single ionized atom, that is deficient one electron, is traveling at in a straight line when it enters a uniform magnetic field of strength 0.750 T. The ion is traveling in the plane of the page, while the magnetic field points into the page. (See the diagram below.) If the mass of the ionized atom is 6.68 x 10-27 kg, what is the frequency in Hz of the resultant circular motion? The charge of a single electron is 1.602 x 10-19 C. a. 7.47 MHz b. 6.33 MHz c. 4.13 MHz d. 3.50 MHz e. .2.86 MHz
A single ionized atom, that is deficient one electron, is traveling at in a straight line when it enters a uniform magnetic field of strength 0.750 T. The ion is traveling in the plane of the page, while the magnetic field points into the page. (See the diagram below.) If the mass of the ionized atom is 6.68 x 10-27 kg, what is the frequency in Hz of the resultant circular motion? The charge of a single electron is 1.602 x 10-19 C. a. 7.47 MHz b. 6.33 MHz c. 4.13 MHz d. 3.50 MHz e. .2.86 MHz
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A single ionized atom, that is deficient one electron, is traveling at in a straight line when it enters a uniform magnetic field of strength 0.750 T. The ion is traveling in the plane of the page, while the magnetic field points into the page. (See the diagram below.) If the mass of the ionized atom is 6.68 x 10-27 kg, what is the frequency in Hz of the resultant circular motion? The charge of a single electron is 1.602 x 10-19 C.
a. 7.47 MHz
b. 6.33 MHz
c. 4.13 MHz
d. 3.50 MHz
e. .2.86 MHz
![### Educational Content on Ionized Atom Motion in a Magnetic Field
#### Problem Statement
A single ionized atom, which is deficient by one electron, travels in a straight line and enters a uniform magnetic field of strength 0.750 T. The ion moves in the plane of the page while the magnetic field points into the page. The scenario is represented by the diagram provided.
- **Mass of the ionized atom:** \(6.68 \times 10^{-27}\) kg
- **Charge of a single electron:** \(1.602 \times 10^{-19}\) C
The inquiry focuses on determining the frequency (in Hz) of the ion's resultant circular motion.
#### Diagram Explanation
- The diagram shows an ion moving in a curved path due to the magnetic field.
- **\(\vec{v}\):** Represents the velocity of the ion, directed to the right initially.
- **\(\vec{F}\):** Represents the magnetic force acting on the ion, perpendicular to the velocity, creating a circular path.
- **\(\vec{B}\):** Indicates the magnetic field direction, denoted by "×" symbols, pointing into the page.
- The dashed blue line indicates the path of circular motion.
#### Answer Choices
- a. 7.47 MHz
- b. 6.33 MHz
- c. 4.13 MHz
- d. 3.50 MHz
- e. 2.86 MHz
#### Solution Approach
To solve this, use the formula for the frequency of a charged particle in a magnetic field:
\[ f = \frac{qB}{2 \pi m} \]
Where:
- \( q \) is the charge of the particle (\(1.602 \times 10^{-19}\) C)
- \( B \) is the magnetic field strength (0.750 T)
- \( m \) is the mass of the particle (\(6.68 \times 10^{-27}\) kg)
Calculating using the given values will lead you to the correct frequency from the provided choices.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ede0637-a731-4aa5-a543-54dacf89e51e%2Fc7e40130-bd4f-4230-a239-b390212086f7%2Fqrujtey_processed.png&w=3840&q=75)
Transcribed Image Text:### Educational Content on Ionized Atom Motion in a Magnetic Field
#### Problem Statement
A single ionized atom, which is deficient by one electron, travels in a straight line and enters a uniform magnetic field of strength 0.750 T. The ion moves in the plane of the page while the magnetic field points into the page. The scenario is represented by the diagram provided.
- **Mass of the ionized atom:** \(6.68 \times 10^{-27}\) kg
- **Charge of a single electron:** \(1.602 \times 10^{-19}\) C
The inquiry focuses on determining the frequency (in Hz) of the ion's resultant circular motion.
#### Diagram Explanation
- The diagram shows an ion moving in a curved path due to the magnetic field.
- **\(\vec{v}\):** Represents the velocity of the ion, directed to the right initially.
- **\(\vec{F}\):** Represents the magnetic force acting on the ion, perpendicular to the velocity, creating a circular path.
- **\(\vec{B}\):** Indicates the magnetic field direction, denoted by "×" symbols, pointing into the page.
- The dashed blue line indicates the path of circular motion.
#### Answer Choices
- a. 7.47 MHz
- b. 6.33 MHz
- c. 4.13 MHz
- d. 3.50 MHz
- e. 2.86 MHz
#### Solution Approach
To solve this, use the formula for the frequency of a charged particle in a magnetic field:
\[ f = \frac{qB}{2 \pi m} \]
Where:
- \( q \) is the charge of the particle (\(1.602 \times 10^{-19}\) C)
- \( B \) is the magnetic field strength (0.750 T)
- \( m \) is the mass of the particle (\(6.68 \times 10^{-27}\) kg)
Calculating using the given values will lead you to the correct frequency from the provided choices.
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