Photoelectron are liberated by altra voilet light of wavelength `3000Å`from a metallic surface for which the photoelectron thershold is `4000 Å` calculate de broglic wavelength of electron with maximum kinetic energy

Explanation

KE = Quantum energy - Theshold energy = `(hc)/(lambda_(1)) - (hc)/(lambda_(2))`
`= (6.626 xx 10^(-34) xx 3 xx 10^(8))/(3000 xx 10^(-10)) - (6.626 xx 10^(-34) xx 3 xx 10^(8))/(4000 xx 10^(-10)) `
`= (6.626 xx 10^(-34) xx 3 xx 10^(8))/(10^(-10)) ((1)/(3000) - (1)/(4000))`
`= 1.656 xx 10^(-19) J`
`1/2mv^(2)=1.656xx10^(-19)`
`rArr m^(2)v^(2) = 2 xx 1.6565 xx 10^(-19) xx 9.1 xx 10^(-31)`
`mv = 5.49 xx 10^(25)`
`rArr lambda = (h)/(mv) = (6.626 xx 10^(-34))/(3.49 xx 10^(-25)) = 1.2 xx 10^(-9) m`