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Question: What will be the velocity of the proton of the uranium when it returns back from \(5m\) away? The in...

What will be the velocity of the proton of the uranium when it returns back from 5m5m away? The initial velocity will be given as 3×105ms13\times {{10}^{5}}m{{s}^{-1}}.
A.1.8×105ms1 B.2.3×105ms1 C.3×105ms1 D.2.7×105ms1 \begin{aligned} & A.1.8\times {{10}^{5}}m{{s}^{-1}} \\\ & B.2.3\times {{10}^{5}}m{{s}^{-1}} \\\ & C.3\times {{10}^{5}}m{{s}^{-1}} \\\ & D.2.7\times {{10}^{5}}m{{s}^{-1}} \\\ \end{aligned}

Explanation

Solution

The law of conservation of energy is the basis for solving this question. According to this conservation, the total energy if the proton reaches towards the uranium will be equivalent to the total energy when it reaches away from uranium. The total energy is given as the sum of the potential energy and the kinetic energy. This all will help you in answering this question.

Complete answer:
Let us assume that the velocity with which the proton reaches back can be VV. The initial velocity of the proton when it moves from the uranium atom has been given as,
U=3×105ms1U=3\times {{10}^{5}}m{{s}^{-1}}
According to the conservation of energy, the total energy if the proton reaches towards the uranium will be equivalent to the total energy when it reaches away from uranium. The total energy is given as the sum of the potential energy and the kinetic energy.
The kinetic energy of the proton when it moves away from the atom will be,
K.E=12mU2K.E=\dfrac{1}{2}m{{U}^{2}}
The potential energy of the proton when it moves away from the atom will be,
PE=Fd1PE=F{{d}_{1}}
The kinetic and potential energies respectively of the proton when it moves towards the atom can be written as,
K.E=12mV2\Rightarrow K.{E}'=\dfrac{1}{2}m{{V}^{2}}
PE=Fd2\Rightarrow P{E}'=F{{d}_{2}}
The total energy will be the sum of these two. That is,
12mU2+Fd1=12mV2+Fd2\Rightarrow \dfrac{1}{2}m{{U}^{2}}+F{{d}_{1}}=\dfrac{1}{2}m{{V}^{2}}+F{{d}_{2}}
Here the distance of the travel will be equal.
d1=d2=5m\Rightarrow {{d}_{1}}={{d}_{2}}=5m
As all other values in the question are equal, the velocities will also be equal.
U=V=3×105ms1\therefore U=V=3\times {{10}^{5}}m{{s}^{-1}}

So, the correct answer is “Option C”.

Note:
The potential energy is the energy associated with a body due to its configuration in space. The kinetic energy is the energy associated with a body due to its motion. The total energy of an isolated system will remain unchanged every time.