When light of 470 nm falls on the surface of potassium metal, electrons are emitted with a velocity of `6.4 xx 10^(4)ms^(-1)`. What is the minimum energy required to remove one moles electrons from potassium metal?

Explanation

kinetic energy of photo electrons `= (1)/(2) mv^(2)`
`= (1)/(2) xx 9.1 xx 10^(-31) xx (6.4 xx 10^(4) )^(2) = 1.864 xx 10^(-21) J`
we know that, Absorbed energy from light = Threshold energy + kinetic energy of photoelectrons
`(hc)/(lambda) = W_(0) +(1)/(2) mv^(2) = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(470 xx 10^(-9))`
`= W_(0) + 1.864 xx 10^(-21) J = W_(0) = 419.206 xx 10^(21) J`.
Energy required to eject one mole electrons
`= 419. 21 xx 10^(-21) xx 6.023 xx 10^(23)`
`= 252.4 xx 10^(3) J mol^(-1) = 252.3 kJ mol^(-1)`