Question

what is the speed of a proton that has been accelerated from rest through a potential difference of -800 v ?

Answer 1

The speed of the proton after being accelerated through a potential difference of -800 V is approximately 1.46 x 10^8 m/s.

Step-by-step explanation:

The speed of a proton can be found using the equation:

Kinetic energy = (1/2)mv^2

Since the proton is accelerated from rest, its initial kinetic energy is zero. The potential difference (-800 V) can be converted to energy in electron volts (eV) using the conversion factor 1 eV = 1.6 x 10^-19 J:

-800 V x (1eV / 1 V) = -800 eV

The proton gains 1 eV of energy for each volt across the gap. Therefore, the proton gains -800 eV of energy. To find the final speed of the proton, we can equate the kinetic energy gained to the kinetic energy equation:

-800 eV = (1/2)mv^2

Solving for v, we can rearrange the equation:

v = sqrt((-1600 eV) / m)

Given the mass of a proton is approximately 1.67 x 10^-27 kg, substituting the values into the equation:

v = sqrt((-1600 eV) / (1.67 x 10^-27 kg))

v ≈ 1.46 x 10^8 m/s

Answer 2

The speed of a proton that has been accelerated from rest through the potential difference is 3.92 x 10⁵ m/s.

How to calculate the speed of the proton?

The speed of a proton that has been accelerated from rest through the potential difference is calculated by applying the following formula as shown below.

K.E = eV

¹/₂mv² = eV

mv² = 2eV

v² = 2eV / m

v = √ (2eV / m)

where;

• V is the magnitude of the potential difference
• e is the charge of proton
• m is the mass of proton

The speed of a proton that has been accelerated from rest through the potential difference is calculated as;

v = √ (2eV / m)

v = √ (2 x 1.6 x 10⁻¹⁹ x 800 / 1.67 x 10⁻²⁷)

v = 3.92 x 10⁵ m/s