Question

Q4. An electron, in a picture tube of a TV set, travelling in a straight line, accelerates uniformly from speed 3x104 to 5x10 m/s along a lenght of 2 cm. (9) a. How much time does the electron spend in this 2 cm region? b. What is the magnitude of the electron's acceleration?

Answer 1

Acceleration a is calculated from equation of motion , " v² = u² + 2 a S "

where v is final speed , u is initial speed and S is the distance moved

a = ( v² - u² ) / ( 2 S ) = { [ ( 500 $\times$ 500 ) - 9 ] $\times$ 10⁸ } / ( 2 $\times$ 2 $\times$ 10^-2 )

we get acceleration a = 6.25 $\times$ 10¹⁴ m/s²

Time t spent by electron in 2 cm distance is calculated using the equation of motion " v = u + a t "

Time t spent by electron in 2 cm distance = ( v - u ) /a = [ ( 500 - 3 ) $\times$ 10⁴ ] / ( 6.25 $\times$ 10¹⁴ )

Time t spent by electron in 2 cm distance = 7.952 $\times$ 10^-9 s = 7.952 ns