In an electron gun, the potential difference between the filament and plate is 3000 V. What will be the velocity of electron emitting from the gun?

Question

In an electron gun, the potential difference between the filament and plate is 3000 V. What will be the velocity of electron emitting from the gun? (A) 3 × 108 m/s  (B) 3.18 × 107 m/s  (C) 3.52 × 107 m/s  (D) 3.26 × 107 m/s

Tardigrade Admin · Accepted Answer

V =3000 volt. (1/2) m v2=e V ⇒ v=(√2 e V/m) ∴ v=√(2 × 1.6 × 1019 × 3000/9.1 × 10-31) 32.6 × 106=3.26 × 107 m / s

Q. In an electron gun, the potential difference between the filament and plate is $3000 V$ . What will be the velocity of electron emitting from the gun?

$3 \times 1 0^{8} m / s$

$3.18 \times 1 0^{7} m / s$

$3.52 \times 1 0^{7} m / s$

$3.26 \times 1 0^{7} m / s$

$V = 3000$ volt.
$\frac{1}{2} m v^{2} = e V$
$\Rightarrow v = \frac{2 e V}{m}$
$∴ v = \frac{2 \times 1.6 \times 1 0 ^{19} \times 3000}{9.1 \times 1 0 ^{- 31}}$
$32.6 \times 1 0^{6} = 3.26 \times 1 0^{7} m / s$

Q. In an electron gun, the potential difference between the filament and plate is 3000V. What will be the velocity of electron emitting from the gun?

3×108m/s

3.18×107m/s

3.52×107m/s

3.26×107m/s