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

Answer 1

it's answer should be option d

Answer 2

Answer:

D) 3.26×10⁷ m/s

Explanation:

Potential difference between filament and plate = V = 3000 volt.  

Whenever a charged particle q is accelerated through a potential difference of V, its kinetic energy increases by an amount of qV, thus velocity can be calculates as -

1/2mv² = eV = v = √2eV/m

Therefore,

v = √2×1.6×10-19×3000/ 9.1×10−31

= 32.6×10-6

= 3.26×10⁷ m/s

Thus, the velocity of electron emitting from the gun will be 3.26×10⁷ m/s