# What is the acceleration of the electron in this region and how long is the electron in the region where it accelerates?An electron in a cathode ray tube of a TV set enters a region where it...

What is the acceleration of the electron in this region and how long is the electron in the region where it accelerates?

An electron in a cathode ray tube of a TV set enters a region where it accelerates uniformly from a speed of 3.5x10^4 m/s to a speed of 1.4x10^6 m/s in a distance of 2.0 cm.

neela | High School Teacher | (Level 3) Valedictorian

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If  the initial speed of the electron is u and the final speed of the electron is v with a constant acceleration a and the distance travelled by the electron is s, then the time t taken by the electron to travel the distance s is given by:

t = distance/average speed.

Or t = s/{(u+v)/2)}.

Or t = 2s/(u+v).

In this case ,u = 3.5*10^4 m/s v = 1.4*10^4 m/s-1 and s = 2cm.

Therefore  t = 2*2cm/(3.5*10^4 m/s+1.4*10^6 m/s).

t = (2*0.02m)/(3.5m/s+140m/s)10^4.

t = (0.04m)/(143.5m/s)10^4.

t =  2.7875*10^(-8) s.

Therefore the time taken by the electron to move 2cm from a speed of 3.5*10^4 m/s to a speed of 1.4 m/s with a uniform acceleration is 2.7875 s approximately.

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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The acceleration of an object can be calculated by the formula:

a = (v^2 - u^2)/2s

where:

u = Initial velocity

v = Final Velocity

s = Distance travelled during increase of velocity from u to v.

Substituting the given values of u, v, and s for the electron in the above equation for distance:

s = [(1.5x10^6)^2 - (3.5x10^4)^2]/[2x(2/100)]

= (2.25x10^12 - 12.25x10^8)/0.04

= (2.25x10^10 - 12.25x10^6)/4

= (2.248775x10^10)/4

= 5621937.5x10^3 m/s^2

The time (t) taken to accelerate is given by:

t = (v - u)/a

Substituting the values of u, v and a in above equation:

t = (1.5x10^6 - 3.5*10^4)/(5621937.5x10^3)

= (1.4658x10^3)/5621937.5

= 0.0002591277 s