A bare helium nucleus has two positive charges and a mass of 6.64 10-27 kg.

(a) Calculate its kinetic energy in joules at 2.00% of the speed of light.


J
(b) What is this in electron volts?


eV
(c) What voltage would be needed to obtain this energy?


V

1.2x10-13

(a) To calculate the kinetic energy of a particle, we can use the formula:

Kinetic energy = (1/2) * mass * velocity^2

Given:
Charge (Q) = 2 positive charges
Mass (m) = 6.64 * 10^-27 kg
Speed (v) = 2% of the speed of light, which is:

v = 0.02 * c

where c is the speed of light (approximately 3 * 10^8 m/s).

Therefore, the velocity of the helium nucleus is:

v = 0.02 * (3 * 10^8 m/s)

Now, we can use this velocity to calculate the kinetic energy. Plugging in the values into the formula:

Kinetic energy = (1/2) * (6.64 * 10^-27 kg) * [(0.02 * (3 * 10^8 m/s))^2]

Calculate this expression to find the answer.

(b) To convert the kinetic energy from joules to electron volts (eV), we can use the conversion factor:

1 eV = 1.6022 * 10^-19 J

Divide the kinetic energy in joules by this conversion factor to get the answer in electron volts.

(c) To find the voltage needed to obtain this energy, we use the equation:

Voltage (V) = Kinetic energy (in joules) / Charge (in coulombs)

In this case, the charge is 2 positive charges. Convert this charge to coulombs by multiplying it by the elementary charge (e), which is approximately 1.6022 * 10^-19 C.

Now, divide the kinetic energy in joules by the charge in coulombs to obtain the voltage needed.