Question

photo electrons are removed with kinetic energy 1.864 into 10 raise to power minus 21 joules when photons of light with energy 4.23 into 10 raise to power minus 19 falls on the metal what is the minimum energy required per mole to remove an electron from potassium metal

Answer 1

First of all E=h*v (v=freq) V=E/h =4.23*10^-19 /6.62*10^-34 =0.63*10^15 =6.3*10^14So KE=hv-hv0 (v=freq) (vo =threshold freq)Substituting known valuesHere we should find hv0 which is the the binding energy [ie) energy required to remove an electron] REWRITING THE ABOVE KE EQUATION TO BINDING ENERGYHvo=hv-KE (hv=6.6*10^-34*6.3*10^14) h is plank constantAnswer will come like 4.082*10^-19That' is binding energy per photon ..So for one mole it is =6.023*10^23 * 4.082*10^-19 (as one mole is equal to 6.023*10^23 photons)=252KJ(approx) which is the required bindind energy to remove one mole of photons
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