Saved Bookmarks
| 1. |
(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 1.76 xx 10^(11) C kg^(-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? |
|
Answer» Solution :(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`. |
|