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Question: With what velocity should a \(\alpha - \)particle travel towards the nucleus of a Cu atom so as to a...

With what velocity should a α−\alpha - particle travel towards the nucleus of a Cu atom so as to arrive at a distance of 10−13m{10^{ - 13}}m.

Explanation

Solution

We can calculate the velocity that is required by α−\alpha - particle travels towards the nucleus of a Cu atom using the atomic number of copper, charge of alpha particle, radius, and mass of alpha particle. We have to remember that the potential energy should be equal to the kinetic energy.

Complete step by step answer:
Given data contains,
Distance at which the α\alpha -particle travels towards the nucleus of a Cu atom is 10−13m{10^{ - 13}}m.
From the nucleus, to arrive at a distance r, the kinetic energy of the alpha particle must be equal to the potential energy of interaction of alpha particle with copper nucleus. This can be given as,
K.E=P.EK.E = P.E
We know that,
K.E=P.EK.E = P.E
12mu2=14πE0×2Ze2r\dfrac{1}{2}m{u^2} = \dfrac{1}{{4\pi {E_0}}} \times \dfrac{{2Z{e^2}}}{r}
Here, m is the mass
V is the velocity
e is the charge of the alpha particle
m is the mass of the alpha particle
Z is the atomic number
r is the distance
π\pi takes the value of 3.143.14 and
E0{E_0} takes the value of 8.85×10−128.85 \times {10^{ - 12}}
The equation 12mu2=14πE0×2Ze2r\dfrac{1}{2}m{u^2} = \dfrac{1}{{4\pi {E_0}}} \times \dfrac{{2Z{e^2}}}{r} is rearranged as,
u2=Ze2πE0×m×r{u^2} = \dfrac{{Z{e^2}}}{{\pi {E_0} \times m \times r}}
Let us now substitute the values of all variables in the equation to get the velocity.
We can calculate the velocity as,
u2=Ze2πE0×m×r{u^2} = \dfrac{{Z{e^2}}}{{\pi {E_0} \times m \times r}}
u2=(29)(1.6×10−19)2(3.14)(8.85×10−12)×(4×1.672×10−27)×10−13{u^2} = \dfrac{{\left( {29} \right){{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{\left( {3.14} \right)\left( {8.85 \times {{10}^{ - 12}}} \right) \times \left( {4 \times 1.672 \times {{10}^{ - 27}}} \right) \times {{10}^{ - 13}}}}
u=6.3×106msec⁡−1u = 6.3 \times {10^6}m{\sec ^{ - 1}}
The velocity at which the α−\alpha - particle travel towards the nucleus of a Cu atom to reach at a distance of 10−13m{10^{ - 13}}m is 6.3×106msec⁡−16.3 \times {10^6}m{\sec ^{ - 1}}.

Note: We have an approach to solve this question,
Alpha particle could approach towards the nucleus of copper upto distance r0{r_0}=10−13m{10^{ - 13}}m
From the nucleus to reach the distance r, the kinetic energy of the alpha particle must be equal to the potential energy of interaction of the alpha particle with the nucleus of copper.
We can give this as,
K.E=P.EK.E = P.E
12mαvα2=kqαqCur0\dfrac{1}{2}{m_\alpha }{v_\alpha }^2 = k\dfrac{{{q_\alpha }{q_{Cu}}}}{{{r_0}}}
We know that alpha particle mass would be 4×1.67×10−27kg4 \times 1.67 \times {10^{ - 27}}kg.
The value of kk is 9×109Nm2/C29 \times {10^9}N{m^2}/{C^2}
We know that charge of an alpha particle is 2×1.6×10−19C2 \times 1.6 \times {10^{ - 19}}C.
We know that charge of copper is given as 29×(1.6×10−19C)29 \times \left( {1.6 \times {{10}^{ - 19}}C} \right)
We know that distance is 10−13m{10^{ - 13}}m
Let us now substitute this values in the expression,
12mαvα2=kqαqCur0\dfrac{1}{2}{m_\alpha }{v_\alpha }^2 = k\dfrac{{{q_\alpha }{q_{Cu}}}}{{{r_0}}}
12(4×1.67×10−27)vα2=(9×109)(2×1.6×10−19)(29×1.6×10−19)10−13\dfrac{1}{2}\left( {4 \times 1.67 \times {{10}^{ - 27}}} \right){v_\alpha }^2 = \left( {9 \times {{10}^9}} \right)\dfrac{{\left( {2 \times 1.6 \times {{10}^{ - 19}}} \right)\left( {29 \times 1.6 \times {{10}^{ - 19}}} \right)}}{{{{10}^{ - 13}}}}
vα2=4000.9×1010{v_\alpha }^2 = 4000.9 \times {10^{10}}
vα=63.25×106m/s{v_\alpha } = 63.25 \times {10^6}m/s
vα=6.325×106m/s{v_\alpha } = 6.325 \times {10^6}m/s
The velocity at which the α−\alpha - particle travel towards the nucleus of a Cu atom to reach at a distance of 10−13m{10^{ - 13}}m is 6.3×106msec⁡−16.3 \times {10^6}m{\sec ^{ - 1}}.