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Galina-37 [17]
3 years ago
14

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
Physics
1 answer:
Nata [24]3 years ago
8 0

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

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