Science, asked by sharfraz8691, 9 months ago

Photons of energy 6 eV are incident on a potassium surface whose work function is 2.1 eV. The corresponding stopping potential required is

Answers

Answered by sumitbiswal560
0

Answer:

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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