Photoelectric Effects and deBroglie Equation: Practice Problem & Solution
When light of wavelength $300 nm$ (nanometer) falls on a photoelectric emitter, photoelectrons are liberated. For another emitter, however, light of $600 nm$ wavelength is sufficient for creating photoemission. What is the ratio of the work functions of the two emitters? (1993)
Solution Explained:
To solve this problem, we apply the core principles of Photoelectric Effects and deBroglie Equation. Understanding the underlying formula is key to arriving at the correct answer below:
The work function is inversely proportional to the threshold wavelength: $\phi = \frac{hc}{\lambda_0}$. The ratio of the work functions is $\frac{\phi_1}{\phi_2} = \frac{\lambda_{02}}{\lambda_{01}} = \frac{600 nm}{300 nm} = \frac{2}{1}$. Thus, the ratio is 2 : 1.
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