Saved Bookmarks
| 1. |
What is the a. momentum, b. speed, and c. de Broglie wavelength of an electron with kinetic energyof 120 eV. |
|
Answer» Solution :KINETIC energy, `K.E = 120 eV = 120 XX 1.6 xx 10^(-19)` `K = K.E = 1.92 xx 10^(-17)J` (a) Momentum of an electron, `P = sqrt(2 mK)` `P = sqrt(2 xx 9.1 xx 10^(-31) xx 1.92 xx 10^(-17))` `P = 5.91 xx 10^(-24) "kg ms"^(-1)` (B) Speed of an electron, `v = (P)/(m) = (5.91 xx 10^(-24))/(9.1 xx 10^(-31)) = 6.5 xx 10^(6) ms^(-1)` (c) de-Broglie wavelength, `lambda = (H)/(P) = (6.6 xx 10^(-34))/(5.91 xx 10^(-24)) = 1.117 xx 10^(-10) = 0.112 xx 10^(-9)m` `lambda = 0.112 m` |
|