InterviewSolution
Saved Bookmarks
| 1. |
The kinetic energy of an electron with de - Broglie wavelength of `0.3 nanometre` isA. `0.168 eV`B. `16.8 eV`C. `1.68 eV`D. `2.5 eV` |
|
Answer» Correct Answer - A `lambda = (h)/(sqrt(2 mE)) rArr E = (h^(2))/( 2 m lambda^(2))` `= ((6.6 xx 10^(-34))^(2))/(2 xx 9.1 xx 10^(-31) xx (0.3 xx 10^(-9))^(2))` `= 2.65 xx 10^(-18) J = 16.8 eV` |
|