# Finding Velocity of Positive ion with Volts given.

• dkr1207
In summary, the problem involves finding the radius of the path of an ion that is accelerated through a potential difference of 250 V and moves directly perpendicular to a magnetic field of 0.50 T. Using the formula for radius, (mass x velocity)/(charge x magnetic field), and the definition of potential difference as energy, the velocity is calculated to be 3.54 x 10^23 cm/s. However, when this value is plugged into the final radius equation, it does not match the known answer of 1.77 cm. After further calculations, the correct answer is obtained.
dkr1207

## Homework Statement

An ion is accelerated through potential diff. of 250 V, directly perpendicular to mag. field.
Find radius of the path of the ion in the field.

Given:
mass (m)=2.5 x 10^-26
charge(q)= 1.6 x 10^-19
Volts= 250 V
Mag. Field (B) = 0.50 T

## Homework Equations

radius= (mass x VELOCITY)/ (charge x B)

## The Attempt at a Solution

I know the answer is 1.77 cm for the radius, but i need to prove how, but i can't seem to find an equation or reason to use Volts to get Velocity in the equation.

What's the definition of "potential difference"? Thinking about that should help you figure out how to use the number of volts. Hint: energy.

Pe

First i find PE with:

PE=Volts x charge

plug in to:

velocity=[(-2/mass) x PE]^(1/2)

seems right...

PE= 250V / 1.6 x 10^-19 C = 1.56 x 10^21 J

plug in:
and you get velocity at 3.54 x 10^23 cm/s

but when you use that number in the final radius equation, its not = to 1.77 cm (and i know that's right)

First i find PE with:

PE=Volts x charge

plug in to:

velocity=[(-2/mass) x PE]^(1/2)

seems right...

PE= 250V / 1.6 x 10^-19 C = 1.56 x 10^21 J

plug in:
and you get velocity at 3.54 x 10^23 cm/s

but when you use that number in the final radius equation, its not = to 1.77 cm (and i know that's right)

nvm... its good now, thanks for the help

## 1. What is the formula for finding velocity of a positive ion with volts given?

The formula for finding velocity of a positive ion with volts given is v = √(2qV/m), where v is the velocity, q is the charge of the ion, V is the voltage, and m is the mass of the ion.

## 2. How do you determine the charge and mass of the positive ion?

The charge and mass of the positive ion can be determined through various methods such as mass spectrometry, ion mobility spectrometry, and charge-to-mass ratio measurements.

## 3. Can the formula be used for all types of positive ions?

Yes, the formula can be used for all types of positive ions as long as the charge and mass values are known.

## 4. What is the unit of measurement for velocity in this formula?

The unit of measurement for velocity in this formula is meters per second (m/s).

## 5. How is the velocity of a positive ion affected by the voltage?

The velocity of a positive ion is directly proportional to the voltage. This means that as the voltage increases, the velocity of the ion also increases.

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