High School

A proton with a mass ([tex]m[/tex]) of [tex]1.67 \times 10^{-27}[/tex] kg, originally traveling with a kinetic energy of [tex]380 \times 10^{-3}[/tex] eV, is decelerated by moving through a potential difference of 122 mV. Find its final speed (in km/s).

Answer :

Final answer:

The final speed of the proton after deceleration through a potential difference of 122 mV is approximately 8.88 x 10^5 km/s.

Explanation:

To find the final speed of the proton, we can use the formula:

V = sqrt(2qV/m) where V is the potential difference, q is the charge of the proton (1.6 x 10^-19 C), and m is the mass of the proton.

Plugging in the given values:

V = sqrt(2 * 1.6 x 10^-19 C * 122 x 10^-3 V / 1.67 x 10^-27 kg)

After calculating, the final speed of the proton is approximately 8.88 x 10^5 km/s.

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