Question

Light with a frequency of 2. 41×10^15 hz ejects electrons from the surface of platinum, which has a work function of 6. 53 ev. What is the minimum de broglie wavelength of the ejected electrons?

Answer 1

The minimum de Broglie wavelength of the ejected electrons = 0.66 nm

The De Broglie wavelength is a wavelength manifested in all of the objects in quantum mechanics which determines the opportunity density of locating the item at a given point of the configuration area. The de Broglie wavelength of a particle is inversely proportional to its momentum.

Frequency = 2. 41×10^15 Hz

Work function =  6. 53 eV = 6. 53 * 1.6 * 10^-19 = 10.45 * 10^-19

ENERGY = Work function +  MAX KINETIC ENERGY

K max = 6.626 * 10^-34 * 2. 41×10^15 - 10.45 * 10^-19

K max = 15.96 *10^-19 - 10.45 * 10^-19

K max = 5.51 * 10^-19

Now, minimum de Broglie wavelength of ejected electron

       λmin= $\frac{h}{\sqrt{2mKmax} }$

               = $\frac{6.626 * 10^-34}{2 * 9.1 * 10^-31*6.25 *10^-19}$

             =  0.66 nm  answer

             

Hence, The minimum de Broglie wavelength of the ejected electrons = 0.66 nm

Learn more about wavelength here:-brainly.com/question/10750459

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