Saved Bookmarks
| 1. |
If photoelectrons are to be emitted from a potassium surface with a speed 6 xx 10^(6) ms^(-1), what frequency of radiation must be used ? (Threshold frequency for potassium is 4.22 xx 10^(14) Hz, h = 6.6 xx 10^(-34) Js, m_(e) = 9.1 xx 10^(-31) kg) |
|
Answer» SOLUTION :Here, `V = 6 xx 10^(6) ms^(-1)` `V_(0) = 4.22 xx 10^(14)` Hz From Einstein.s photoelectric equation, `K.E = (1)/(2) mv^(2) = H(upsilon - upsilon_(0))` `upsilon = (1)/(2) (mv^(2))/(h) + upsilon_(0)` `= (1)/(2) xx (9.1 xx 10^(-31) + (6 xx 10^(6))^(2))/(6.6 xx 10^(-34)) + 4.22 xx 10^(14)` `= (2.48 xx 10^(14)) + (4.22 xx 10^(14))` `upsilon = 6.7 xx 10^(14)` Hz |
|