If barium has a work function of 2.60 eV,what is its cutoff wavelength when used as a phototube target? (h = 6.63 x 10^-34J*s,c = 3.00 x 10⁸ m/s,1 eV = 1.60 x 10^-19 J and 1 nm = 10^-9 m) A)398 nm B)478 nm C)497 nm D)596 nm

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Quizplus · Accepted Answer

The cutoff wavelength is calculated using the formula λ = hc/Φ, where Φ is the work function. Converting the work function from eV to Joules (2.60 eV = 2.60 x 1.60 x 10^-19 J), and using h = 6.63 x 10^-34 J·s and c = 3.00 x 10^8 m/s, we find λ = (6.63 x 10^-34 J·s * 3.00 x 10^8 m/s) / (2.60 x 1.60 x 10^-19 J) = 478 nm.

If Barium Has a Work Function of 2