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Question: With what velocity, (\(n \times {10^6}m{s^{ - 1}}\)) should an \(\alpha\)-particle travel towards th...

With what velocity, (n×106ms−1n \times {10^6}m{s^{ - 1}}) should an α\alpha-particle travel towards the nucleus of a copper atom so as to arrive at a distance 10−13m{10^{ - 13}}m from the nucleus of the copper atom? Round off to the nearest integer.

Explanation

Solution

The velocity of the alpha particles moving towards the nucleus of the copper atom can be calculated by using the relation where the kinetic energy of the alpha particle is equal to the potential energy of the alpha particle.

Complete step by step answer:
Given,
The distance at which the α\alpha-particle travel towards the nucleus of a copper atom is 10−13m{10^{ - 13}}m.
For the nucleus to arrive at the distance r, the Kinetic energy should be equal to the potential energy.
It is written as shown below
K.E = P.E
So,
12mυ2=14πE0×2Ze2r\dfrac{1}{2}m{\upsilon ^2} = \dfrac{1}{{4\pi {E_0}}} \times \dfrac{{2Z{e^2}}}{r}
Where,
m is the mass
υ\upsilon is the velocity
e is the charge of the alpha particle
Z is the atomic mass
r is the radius
The above equation is rearranged as shown below.
υ2=Ze2πE0×m×r{\upsilon ^2} = \dfrac{{Z{e^2}}}{{\pi {E_0} \times m \times r}}
The value of π\pi is 3.14
The value of E0{E_0} is 8.85×10−128.85 \times {10^{ - 12}}
The atomic mass of copper is 29.
The value of charge of alpha particle is 1.6×10−191.6 \times {10^{ - 19}}
To calculate the velocity, substitute the values in the above equation.
⇒υ2=(29)(1.6×10−19)2(3.14)(8.85×10−12)×(4×1.672×10−27)\Rightarrow {\upsilon ^2} = \dfrac{{(29){{(1.6 \times {{10}^{ - 19}})}^2}}}{{(3.14)(8.85 \times {{10}^{ - 12}}) \times (4 \times 1.672 \times {{10}^{ - 27}})}}
⇒υ=6.3×106msec⁡−1\Rightarrow \upsilon = 6.3 \times {10^6}m{\sec ^{ - 1}}
⇒υ=6×106msec⁡−1\Rightarrow \upsilon = 6 \times {10^6}m{\sec ^{ - 1}}
Thus, with 6×1066 \times {10^6} ms−1m{s^{ - 1}} velocity an α\alpha-particle travel towards the nucleus of a copper atom so as to arrive at a distance 10−13m{10^{ - 13}}m from the nucleus of the copper atom.

Note:
Make sure to round off the calculated value of velocity of the alpha particles to its nearest integer value which means that the value should be not written in decimal form. The alpha particle is the 24He2+_2^4H{e^{2 + }} ion which when moved towards the nucleus of the copper atom, it is repelled by the nucleus of the copper atom.