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The photoelectric work function of potassium is 2.3 eV. If light having wavelength of 2800 A^@ falls on potassium , find the kinetic energy in electron volts of the most energetic electrons ejected. |
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Answer» SOLUTION :Given W = `2.3 E V, lambda = 2800 A^@` `therefore` E ("in" e V)= (12375)/(lambda ("in" A^@)) = (12375)/(2800) = 4.4 e V` `k_(max) = E-W = (4.4 - 2.3 ) eV = 2.1 eV` |
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