The energy that should be added to an electron, to reduce its deBroglie wavelength from  2×10−9 m to 0.5×10−9 m will be______(in eV) (h=6.6×10−34 Js ,  me=9×10−31 kg)

Given datah= Plank’s constant =6.6×10−3 Jsme= Mass of electron= 9×10−31 kgWe use De-broglie wavelength λ=hp=h2mE∵E=p22m ⇒E=h22mλ2 Thus we have   ΔE=h22m1λ12−1λ22Using λ1=0.5×10−9 m and  λ2=2×10−9 mΔE=6.6×10−3422×9×10−3110.5×10−92−12×10−92ΔE=2.42×10−37[4−0.25]×1018=9.075×10−19J=9.075×10−191.6×10−19eV∵1ev=1.6×10−19J∴ΔE=5.67eV