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n electron gun accelerates an electron across a potential difference of $200$200​ V. Take the mass of an electron to be $9.11\times10^{−31}$9.11×10−31​ kg.Calculate the final speed of the electro

Question

n electron gun accelerates an electron across a potential difference of 200200200​ V. Take the mass of an electron to be 9.11×10−319.11\times10^{−31}9.11×10−31​ kg.Calculate the final speed of the electro

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Solution

The kinetic energy (KE) of the electron is given by the equation KE = 1/2 mv^2, where m is the mass of the electron and v is its velocity.

The electron gains kinetic energy from the electric potential difference (V), and the energy gained is given by the equation KE = eV, where e is the charge of the electron.

Setting these two equations equal to each other, we get:

1/2 mv^2 = eV

We can solve for v:

v = sqrt(2eV/m)

Substituting the given values:

v = sqrt(2 * 1.610^-19 C * 200 V / 9.1110^-31 kg)

v = sqrt(6.410^-17 J / 9.1110^-31 kg)

v = sqrt(7.02*10^13 m^2/s^2)

v = 2.65*10^7 m/s

So, the final speed of the electron is approximately 2.65*10^7 m/s.

This problem has been solved

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