(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) Use `eV=(mv^(2)//2)" i.e., "v=[(2eV//m)]^(1//2),v=1.33xx10^(7)ms^(-1)`
(b) If we use the same formula with `V = 10^(7) V`, we get `v = 1.88 × 10^(9) m s^(-1)`.
This is clearly wrong, since nothing can move with a speed greater than the speed of light `(c = 3 xx10^(8) m s^(-1))`. Actually, the above formula for kinetic energy `(m v^(2)//2)` is valid only when `(v//c) lt lt 1`. At very high speeds when (v/c ) is comparable to (though always less than) 1, we come to the relativistic domain
where the following formulae are valid:
Relativistic momentum `p = m v`
Total energy `E = m c^(2)`
Kinetic energy `K = m c^(2) - m_(0) c^(2)` ,
where the relativistic mass m is given by `m=m_(0)(1-(v^(2))/(c^(2)))^(-1//2)`
`m_(0)`is called the rest mass of the particle. These relations also imply:
`E=(p^(2)c^(2)+m_(0)^(2)c^(4))^(1//2)`
Note that in the relativisitc domain when v/c is comparable to 1, K or energy `ge m_(0)c^(2)` (rest mass energy). The rest mass energy of electron is about 0.51 Mev. Thus a kinetic energy of 10 Mev, being much greater than electron’s rest mass energy, implies relativistic domain. Using relativistic formulas, v (for 10 Mev kinetic energy) `= 0.999 c`.