Photoelectrons are liberated by ultraviolet light of wavelength 3000 \AA from a metallic surface for — Atomic Structure Chemistry Question
Question
Photoelectrons are liberated by ultraviolet light of wavelength 3000 \AA from a metallic surface for which the photoelectric threshold is 4000 \AA. The de-Broglie wavelength of electrons emitted with maximum kinetic energy is
Answer: C
💡 Solution & Explanation
K.E. $= \frac{12400}{3000} - \frac{12400}{4000} = 4.133 - 3.1 = 1.033$ eV. The de-Broglie wavelength $\lambda = \frac{12.27}{\sqrt{K}} \text{ \AA} = \frac{12.27}{\sqrt{1.033}} \text{ \AA} \approx 12.07$ \AA. Therefore, correct answer is C.
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