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

(a) Calculate the time of flight for both electron and proton
(b) Suppose if a neutron is allowed to fall what is the time of flight ?
(c ) Among the three which one will reach the bottom first?
( Take `m_(p) = 1.6xx10^(-27)kg , m_(c) = 9.1 xx10^(-31) " kg and g " = 10 ms^(-2))`

Explanation

(a) Time of flight for electron
`t_e = sqrt((2hm_e)/(eE))`
`h = 1 xx 10^-3 m,m_e = 9.1 xx 10^-31 kg`
`g = 10ms^-2`
`e = 1.6 xx 10^-19C`
`E = V/d = 5/(1 xx 10^-3) = 5000Vm^-1`
`t_e = sqrt((2 xx 1 xx10^-3xx9.1xx10^-31)/(1.6xx10^-19xx5000)`
`= sqrt((18.2xx10^-34)/(8xx10^-16)) = sqrt(2.275xx10^-18)`
`1.5xx10^-9s approx1.5ns.`
Time of flight for proton
`t_P = sqrt((2hm_r)/(eE))`
`h = 1xx10^-3m,m_P = 1.6xx10^-27kg`
`e=1.6xx10^-19,E=5000Vm^-1`
`t_P=sqrt((2xx1xx10^-3xx1.6xx10^-27)/(1.6xx10^-19xx5000))`
`= sqrt((3.2xx10^-30)/(8xx10^-16)) =sqrt(3.2/8xx10^-14)`
`sqrt(0.4xx10^-14)`
`=sqrt(0.4xx10^4xx10^-14xx10^-14)`
`=sqrt(40000xx10^-18)approx63xx10^-9s`
`approx63ns.`
(b) Time of flight for neutron
`t_n = sqrt((2h)/g)`
`t_n = sqrt((2xx1xx10^-3)/10)`
`= sqrt(2xx10^-4)`
`= 1.414xx10^-2`
`= 14.14xx10^-3`
`t_n = 14.14ns.`
From the values of time of flight for electron, proton and neutron it is found that time of flight for eletron is the least.
Hence electron will reach first.