The acceleration of an electron in an electric field of magnitude 50 V/cm,

The acceleration of an electron in an electric field of magnitude 50 V/cm, if e/m value of the electron is 1.76×10¹¹ C/kg will be

Options

(a) 8.8×10¹⁴ m/sec²
(b) 3×10¹³ m/sec²
(c) 5.4×10¹² m/sec²
(d) Zero

Correct Answer:

8.8×10¹⁴ m/sec²

Explanation:

Acceleration = a = eE / m

⇒ a = 1.76 × 10¹¹ × 50 × 10 ¹²

⇒ a = 8.8 × 10 ¹⁴ m/ sec²

admin:

View Comments (1)

  • E=50v/cm
    =5000v/m
    e/m=1.76×10power-11
    F=ma
    F=qĒ
    ma=qĒ
    a= e/m×Ē
    =1.76×10 power11 × 5000
    =8.8×10 power 14 m/sec sq.

Related Questions

  1. 10 mA current can pass through a galvanometer of resistance 25 ohm.
  2. In insulators (CB is Conduction Band and VB is Valence Band)
  3. In an LCR circuit, capacitance is changed from C to 2C. For the resonant frequency
  4. In which of the follows decays, the element does not change?
  5. When two displacements represented by y₁=a sin(?t) are superimposed the motion is