If barium has a work function of 2.6 eV,what is its cutoff wavelength when used as a phototube target? (h = 6.63 × 10−³⁴ J⋅s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10−¹⁹ J and 1 nm = 10−⁹ m) A)765 nm B)478 nm C)777 nm D)380 nm

Question

Quizplus · Accepted Answer

The cutoff wavelength can be calculated using the formula:

λ = hc/Φ

where λ is the cutoff wavelength, h is Planck's constant, c is the speed of light, and Φ is the work function.

Substituting the given values:

λ = [(6.63 × 10^-34 J·s) x (3.00 × 10^8 m/s)] / (2.6 eV x 1.60 × 10^-19 J/eV)
λ = 4.78 x 10^-7 m = 478 nm

Therefore, the cutoff wavelength of barium as a phototube target is 478 nm, which corresponds to choice B.

If Barium Has a Work Function of 2