Question

if electrons are accelerated from rest in a 30 mev tandem van de graaff accelerator, what is their final kinetic energy?

Answer 1

The final kinetic energy of an electron accelerated from rest in a 30 MeV tandem Van de Graaff accelerator is -4.8 x 10^-13 J.

Step-by-step explanation:

The final kinetic energy of an electron accelerated from rest in a 30 MeV tandem Van de Graaff accelerator can be calculated by converting the potential energy gained into kinetic energy. The potential energy gained by the electron can be calculated using the equation PE = qeV, where qe is the electron charge and V is the voltage. Since the electron is accelerated through a voltage of 30 MeV, which is equivalent to 30 million volts, we can use the charge of an electron (qe = -1.6 x 10^-19 C) to find the potential energy gained. The final kinetic energy is then equal to the calculated potential energy.

Calculating the potential energy gained:

PE = (1.6 x 10^-19 C)(30 x 10^6 V) = -4.8 x 10^-13 J

Therefore, the final kinetic energy of the electron is -4.8 x 10^-13 J.

Answer 2

The final kinetic energy of the electron accelerated through a voltage of 30 MeV is 48 x 10⁻¹³ J.

Step-by-step explanation:

When an electron is accelerated from rest through a voltage of 30 MeV, its final kinetic energy can be calculated using the formula K = qV, where K is the kinetic energy, q is the charge of the electron, and V is the voltage. Since the charge of an electron is 1.6 x 10⁻¹⁹ C, we can calculate the final kinetic energy as:

K = (1.6 x 10⁻¹⁹ C) x (30 x 10⁶ V) = 48 x 10⁻¹³ J

Therefore, the final kinetic energy of the electron is 48 x 10⁻¹³ J.