(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter.Ignore the small initial speeds of the electrons.The specific charge of the electron ,i.e., its
`(e)/(m)` is given to be `1.76xx10^(11)Ckg^(-1)`
(b) Use the same formula you employ in (a)to obtain electron speed for an collector potential of 10 MV.Do you see what is wrong ?in what way is the formula to be modified?

Explanation

Here `V=500 V,(e )/(m)=1.76xx10^(10)(C )/(kg)`
`m_(0)=9.1xx10^(-31)` kg
(a) Kinetic energy of electron,
`(1)/(2)mv^(2)=eV`
`therefore v=sqrt((2eV)/(m))`…….(1)
`therefore=sqrt((2xx1.76xx10^(11)xx500)/(1))`
`therefore v=sqrt(1.76xx10^(14))`
`therefore v=1.3266xx10^(7)`
`therefore v~~1.33xx10^(7)m//s`
(b)`V.=10 MV=10xx10^(6)=10^(7)V`
`therefore` From equation (1),
`v.=sqrt((2eV)/(m))=sqrt(2xx1.76xx10^(11)xx10^(7))`
`=sqrt(3.52xx10^(10))`
`=1.876xx10^(9)`
`~~1.88xx10^(9)m//s`
Such speed is not possible.Because speed of light is `|c=3xx10^(8)ms^(-1)|`.Any particle can not have velocity more than speed of light .So above equation can only be used for speed `vltc`
`therefore K=mc^(2)-m_(0)c^(2)=(m-m_(0))c^(2)`
Where `m_(0)`=rest mass of particle
and `m=(m_(0))/(sqrt(1-v^(2)/c^(2)))`
`therefore eV=(m_(0)c^(2))/(sqrt(1-(v^(2))/(c^(2))))-m_(0)c^(2)[because K=eV]`
`therefore (eV)/(m_(0)c^(2))=(1)/(sqrt(1-(v^(2))/(c^(2))))-1`
`therefore(eV)/(m_(0)C^(2))+1=(1)/(sqrt(1-v^(2)/(c^(2))))`
`therefore=(1.6xx10^(-19)xx10^(7))/(9.1xx10^(-31xx9xx10^(16)))+1=(1)/sqrt(1-(v^(2))/(c^(2)))`
`19.54+1=(1)/(sqrt(1-(v^(2))/(c^(2)))`
`1-(v^(2))/(c^(2))=(1)/((20.54)^(2))`
`therefore1-(1)/((20.54)^(2))=(v^(2))/(c^(2))`
`therefore1-0.00237=(v^(2))/(c^(2))`
`0.99763=(v^(2))/(c^(2))`
`therefore sqrt(0.99763)=(v)/(c )`
`therefore` v=0.999c