Question

An electron and a proton are allowed to fall through the separation between the plates of a parallel plate capacitor of voltage 5 V and separation distance h = 1 mm as shown in the figure.

1. Calculate the time of flight for both electron and proton.
2. Suppose if a neutron is allowed to fall, what is the time of flight?
3. Among the three, which one will reach the bottom first?

(Take m_p = 1.6 x 10^-27 kg, m_e= 9.1 x 10^-31 kg and g = 10 m s^-2)

Answer 1

Potential difference between the parallel plates V = 5 V

Separation distance, h = 1 mm =1 x 10^-3 m

Mass of proton, mp = 1.6 x 10^-27 kg

Mass of proton, m =9.1 x 10^-31 kg

Charge of an a proton (or) electron, e— 1.6 x 10^-19 C
[u = 0; s = h]

From equation of motion, S = ut +`1/2 "at"^2`

From equation of motion, h = `1/2 "at"^2`

t = `sqrt("2h"/"a")`

Acceleration of an electron due to electric field, a = `"F"/"m" = "eE"/"m"   ["E" = "V"/"d"]`

(a) Time of flight for both electron and proton,

`"t"_"e" = sqrt((2"hm"_"e")/"eM")`

`= sqrt((2 xx 1 xx 10^-13 xx 9.1 xx 10^-31 xx 10^3)/(1.6 xx 10^-19 xx 5))`

`= sqrt((18.2 xx 10^-34 xx 10^-3)/(8 xx 10^-19))`

`= sqrt(2.275 xx 10^-15 xx 10^-3)`

`"t"_"e" = 1.5 xx 10^-9`s

`"t"_"e" = 1.5` ns     ....(1)

`"t"_"p" = sqrt((2"hm"_"e")/("eM")) = sqrt((2"hm"_"p" * "d")/("eV"))`

`= sqrt((2 xx 1 xx 10^-3 xx 1.6 xx 10^-27 xx 10^-3)/(1.6 xx 10^-19 xx 5))`

`= sqrt((2 xx 10^-33)/(5 xx 10^-19))`

`= sqrt(0.4 xx 10^-14) = 6.32 xx 10^-8`

`"t"_"p" = 63 xx 10^-9`

t_p = 63 ns……. (2)

(b) time of flight of neutron

t_n = `sqrt("2h"/"g") = sqrt((2 xx 1 xx 10^3)/10)`

`= sqrt(0.2 xx 10^-3)`

t_n = 0.0141 s = 14.1 x 10^-3 s

t_n = 14.1 x 10^-3 ms ……. (3)

(c) Compairision of values 1,2 and 3. The electron will reach the bottom first.