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

Explanation

(a) Given , voltage , V = 500 V
Specific charge , e/m =`1.76xx10^(11) C kg^(-1)`
We know that
We know that
`eV=1/2mv^2`
`impliesv=((2eV)/m)^(1//2)`
`v=((2xx1.6xx10^(-19)xx500)/(9.1xx10^(-31)))^(1//2)`
`v=1.33xx10^7ms^(-1)`
This is the required speed of emitted electrons.
(b) If we use the same formula with `V=10^7` volt, we get `v=1.88xx10^9ms^(-1)` . This is clearly incorrect , since nothing can move with a speed greater than the speed of light `(c=3xx10^8ms^(-1))` . Actually , the above formula for kinetic energy `(1/2mv^2)` is valid only when `v/cltlt1`. At very high seeds when it is comparable to speed of light ,we come to the relativistic domain where Relativistic momentum,
p = mv
Total energy , `E=mc^2`
Kinetic energy , `E_k=mc^2-m_0c^2`.
`mc^2=0.51+10=10.51MeV`
`m_0/sqrt(1-v^2/c^2)c^2=10.51`
where `m_0` is the rest mass
`1-v^2/c^2=(m_0^2c^4)/(10.51xx10.51)`
`=(0.51xx0.51)/((10.51)^2)`
`impliesv^2/c^2=0.9976`
or v = 0.999 c