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Question: Under the action of a given columbic force the acceleration of an electron is \(2 \cdot 2 \times {10...

Under the action of a given columbic force the acceleration of an electron is 2⋅2×1022ms−22 \cdot 2 \times {10^{22}}m{s^{ - 2}} .Then the magnitude of the acceleration of a proton under the action of same force is nearly.
A) 1⋅6×10−19ms−21 \cdot 6 \times {10^{ - 19}}m{s^{ - 2}}.
B) 9⋅1×1031ms−29 \cdot 1 \times {10^{31}}m{s^{ - 2}}.
C) 1⋅5×1019ms−21 \cdot 5 \times {10^{19}}m{s^{ - 2}}.
D) 0⋅6×1027ms−20 \cdot 6 \times {10^{27}}m{s^{ - 2}}.

Explanation

Solution

The force is defined as the product of mass and acceleration, the unit of the force is equal to kgms−2kgm{s^{ - 2}} but it is taken in newtons i.e. N. The force is also known as push or pull and also it can be used to change the motion of any object.

Formula used: The formula of the applied force is given by,
⇒F=ma\Rightarrow F = ma
Where force is F, the mass is m and acceleration is a.

Complete step by step solution:
It is given in the problem that the acceleration of the electron is 2⋅2×1022ms−22 \cdot 2 \times {10^{22}}m{s^{ - 2}} under the action of columbic force and we need to find the acceleration of the proton under the same force. The acceleration of a proton can be calculated by applying the formula of force.
The formula of the applied force is given by,
⇒F=ma\Rightarrow F = ma
Where force is F, the mass is m and acceleration is a.
Since the force on the proton and electron is equal therefore we get,
⇒m1a1=m2a2\Rightarrow {m_1}{a_1} = {m_2}{a_2}

Where mass of electron ism1{m_1}, the acceleration of electron isa1{a_1}, the mass of proton is m2{m_2} and acceleration of the proton is a2{a_2}.
⇒m1a1=m2a2\Rightarrow {m_1}{a_1} = {m_2}{a_2}
Mass of electron ism1=9⋅1×10−31kg{m_1} = 9 \cdot 1 \times {10^{ - 31}}kg, the mass of proton is m2=1⋅6×10−27kg{m_2} = 1 \cdot 6 \times {10^{ - 27}}kg and acceleration of electron isa1=2⋅2×1022ms−2{a_1} = 2 \cdot 2 \times {10^{22}}m{s^{ - 2}}.
⇒9⋅1×10−31×2⋅2×1022=1⋅6×10−27×a2\Rightarrow 9 \cdot 1 \times {10^{ - 31}} \times 2 \cdot 2 \times {10^{22}} = 1 \cdot 6 \times {10^{ - 27}} \times {a_2}
⇒a2=9⋅1×10−31×2⋅2×10221⋅6×10−27\Rightarrow {a_2} = \dfrac{{9 \cdot 1 \times {{10}^{ - 31}} \times 2 \cdot 2 \times {{10}^{22}}}}{{1 \cdot 6 \times {{10}^{ - 27}}}}
⇒a2=20⋅02×10−91⋅6×10−27\Rightarrow {a_2} = \dfrac{{20 \cdot 02 \times {{10}^{ - 9}}}}{{1 \cdot 6 \times {{10}^{ - 27}}}}
⇒a2=12⋅5×1018ms−2\Rightarrow {a_2} = 12 \cdot 5 \times {10^{18}}m{s^{ - 2}}.
⇒a2=1⋅3×1019ms−2\Rightarrow {a_2} = 1 \cdot 3 \times {10^{19}}m{s^{ - 2}}.
∴a2≈1⋅5×1019ms−2\therefore {a_2} \approx 1 \cdot 5 \times {10^{19}}m{s^{ - 2}}.
The acceleration of the proton under the same force is equal toa2=1⋅5×1019ms−2{a_2} = 1 \cdot 5 \times {10^{19}}m{s^{ - 2}}. The correct answer for this problem is option C.

Note: The students are advised to understand and remember the formula of the force as it is very important in solving the problem like these. The students are also advised to remember the mass of the electron and proton in kilograms as sometimes is not given in the problem.