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

Velocity of emitted electrons
`= 6.4 xx 10^(4) ms^(-1)`
Kinetic energy of emitted electrons K.E. `= 1/2 mv^2`
`=1/2 xx 9.1 xx 10^(-31) xx (6.4 xx 10^4)^2`
`:. = 1.864 xx 10^(-21) kg m^2 s^(-2)`
`= 1.864 xx 10^(-21) J`
Energy of photon,
`E = hv = (hc)/(lambda)`
`E = (6.63 xx 10^(-34) xx 3.0 xx 10^8)/(470 xx 10^(-9))`
`=4.23 xx 10^(-19) J`
Minimum energy required per mole
`= 421.14 xx 10^(-21) xx 6.023 xx 10^(23)`
`= 253.6 xx 10^3 J mol^(-1)`
or `=253.6 kJ mol^(-1)`.