Answer:
v = 1.45 × 10⁷ m/s
Step-by-step explanation:
Given:
Inner radius of the cylinder, r₁ = 20 mm = 0.2 m
outer radius of the cylinder, r₂ = 80 mm = 0.8 m
Potential difference, ΔV = 600V
Now, the work done (W) in bringing the charge in to the inner conductor
W = [tex]\frac{1}{2}mv^2[/tex]
where, m is the mass of the electron = 9.1 × 10⁻³¹ kg
v is the velocity of the electron
also,
W = qΔV
where,
q is the charge of the electron = 1.6 × 10⁻¹⁹ C
equating the values of work done and substituting the respective values
we get,
qΔV = [tex]\frac{1}{2}mv^2[/tex]
or
1.6 × 10⁻¹⁹ × 600 = [tex]\frac{1}{2}\times 9.1\times 10^{-31}v^2[/tex]
or
[tex]v = \sqrt\frac{2\times 600\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}[/tex]
or
v = 14525460.78 m/s
or
v = 1.45 × 10⁷ m/s
