Photons of energy 6 eV are incident on a potassium surface whose work function is 2.1 eV. The corresponding stopping potential required is
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From photo-electric equation, eV
From photo-electric equation, eV 0
From photo-electric equation, eV 0
From photo-electric equation, eV 0 = E−φ
From photo-electric equation, eV 0 = E−φ eV
From photo-electric equation, eV 0 = E−φ eV 0
From photo-electric equation, eV 0 = E−φ eV 0
From photo-electric equation, eV 0 = E−φ eV 0 =(6−2.1)eV
From photo-electric equation, eV 0 = E−φ eV 0 =(6−2.1)eVV
From photo-electric equation, eV 0 = E−φ eV 0 =(6−2.1)eVV 0
From photo-electric equation, eV 0 = E−φ eV 0 =(6−2.1)eVV 0
From photo-electric equation, eV 0 = E−φ eV 0 =(6−2.1)eVV 0 = 3.9 V
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