Question

In a photoelectric experiment, with light of wavelength `lambda`, the fastest election has speed v. If the exciting wavelength is changed to `(3lambda)/4`, the speed of the fastest emitted electron will become
A. `= v((4)/(3))^(1/2)`
B. `= v((3)/(4))^(1/2)`
C. `gt v((4)/(3))^(1/2)`
D. `lt v((4)/(3))^(1/2)`

Answer 1

Correct Answer - C
`hv_(0)^(2) - hv_(0) = (1)/(2) m v^(2)`
`:. (4)/(3) hv_(0) - hv_(0) = (1)/(2) m v^(-2)`
`:. (v^(2))/(v^(2) = ((4)/(3) v - v_(0)))/(v - v_(0)) :. V ^(1) = v sqer(((4)/(3) v - v_(0)))/(v - v_(0)))`
`:. V^(1) gt sqrt((4)/(3))`