Question

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In a demonstration of the photoelectric effect, suppose that a minimum energy of 5*10^-19J (3.1 eV) is required to dislodge an electron from a metal surface. What is the wavelength (in nm) of radiation for which the detector registers a response? (Planks constant h=6.63*10^-34 kgm^2/s)

Answer 1

$$E=\frac{hc}{\lambda}$$

$$5\cdot 10^{-19}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^{8})}{\lambda}$$

$$\lambda=4\cdot 10^{-7}m=400\ nm$$