Light with a wavelength 310 nm fell on strontium surface, the electrons were ejected. If maximum kinetic energy of an ejected electron is 1.5 eV. Then [Given : lambda_(e) = sqrt((150)/(DeltaV)) Å where Delta V= Voltage difference of battery]

Light with a wavelength 310 nm fell on strontium surface, the electron

1.	Light with a wavelength 310 nm fell on strontium surface, the electrons were ejected. If maximum kinetic energy of an ejected electron is 1.5 eV. Then [Given : lambda_(e) = sqrt((150)/(DeltaV)) Å where Delta V= Voltage difference of battery]
Answer» de-Broglie wavelength of electron is `10 Å` Work fuction of strontium is `2.5 eV` Threshold wavelength for strontium metal will `496 nm` All EJECTED phot ELECTRONS will have kinetic energy `= 1.5 eV` Solution :`Delta E = (1240)/(310 nm) eV = 4.0 eV` `(KE_("max"))_(e) = 1.5 eV = q Delta V` `Delta V = 1.5 V` (A) `lambda_(e) = sqrt((150)/(Delta V))Å = sqrt((150)/(1.5)) Å = 10 Å` (B) `Delta E = KE_(e) + w.f` `3 = 1.5 + w.f.` WF `= 2.5 eV` (C) `lambda = (1240)/(w.f.) nm = (1240)/(2.5) nm = 496 nm`