Question

A bare helium nucleus consists of two protons and two neutrons
and has a mass of 6.64 × 10^{–27} kg

Calculate its kinetic energy in joules at 2.00% of the speed of light. Use the non-relativistic formula.

What is this in electron volts?

What voltage would be needed to obtain this energy?

Answer #1

The non relativistic formula for kinetic energy is given as,

K.E = 1/2×m×V^2

So here V = 0.02 × velocity of light

M = 6.64 ×10^-27 kg

So on plugging the value we will get,

K.E = 0.5 ×6.64 ×10^-27×(0.02*3*10^8)^2

Therefore K.E = 1.1952 × 10-13 J

Now in eV we have the same relationship as

K.E = 745985.197 eV

Now the voltage that would be needed for this energy is obtained by the formula as,

K.E = q×V

So V = k.E/q

Here total charge we will take charge of two proton as they are present here and no charge on neutron so that,

q = 2× charge of proton

So that , V = 745985.197/2×1.602×10^-19

Therefore V = 2.3284 × 1024 Volt which is required answer.

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