Question

If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed? (e = 1.60 × 10-19 C, k = 1/4πε0 = 8.99 × 109 N · m2/C2, m el = 9.11 x 10-31 kg) If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed? ( = 1.60 × 10-19 C, = 1/4π = 8.99 × 109 N · m2/C2, el = 9.11 x 10-31 kg) 4.9 × 107 m/s 3.9 × 107 m/s 2.9 × 107 m/s 5.9 × 107 m/s

Answer 1

Answer:

So speed of electron will be $5.89\times 10^{-7}m/sec$

Explanation:

We have given potential difference V = 9.9 KV

Charge on electron $e=1.6\times 10^{-19}$

So energy of electron $E=eV=1.6\times 10^{-19}\times 9.9\times 10^3=15.84\times 10^{-16}J$

This energy of electron will be equal to kinetic energy of electron

So $\frac{1}{2}mv^2=15.84\times 10^{-16}J$

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=15.84\times 10^{-16}$

$v=5.89\times 10^{-7}m/sec$

So speed of electron will be $5.89\times 10^{-7}m/sec$