Answer :
Answer:5101.35v
Explanation:
Radius of gold nucleus=7.3×10-15m and a charge of +79e
Q= 79e
e=1.6×10^-19
q= +2e
The nucleus is considered as the point charge where the potential energy between the charges are
U = 1/(4×3.142×Eo)×(qQ)/r
Where r is distance between the charges and the nucleus
r=R+d
V=U/q
U= 1/(4×3.142×Eo)×Q/r
V= 1/(4×3.142×Eo)×Q/(r+d)
9.0×10^9 ×(79×10^-19)/(7.3×10^-15)+(1.5×10^-14)
V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)
V= 9×10^9×(5.67×10^-14)= 5101.35v
A potential difference of 7584 V is required to accelerate an α-particle so that it has just enough energy to reach a distance of 1.5×10⁻¹⁴mm from the surface of a gold nucleus.
Potential difference:
The gold nucleus has a radius of R = 7.3×10⁻¹⁵m and a charge of Q = +79e
The alpha particle has a charge of q = +2e
The electric potential energy is given by:
[tex]U=\frac{1}{4\pi\epsilon_o}\frac{qQ}{R+d}[/tex]
where d is the distance between the alpha particle and the nucleus of the gold atom = 1.5×10⁻¹⁴m
Potential V is defined as:
V = U/q
[tex]V=\frac{1}{4\pi\epsilon_o}\frac{Q}{R+d}[/tex]
[tex]V=\frac{9.0\times10^9 \times(79\times1.6\times10^{-19})}{(7.3\times10^{-15}+1.5\times10^{-14})}[/tex]
V = 5101 V
So, 5101 Volts of potential difference is required for the alpha particle.
Learn more about electric potential:
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