The work function of tungsten is 4.50 , eV. T | vedclass

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Question

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE_{\max})$ of the emitted electron is given by:$KE_{\max} = E - \phi_

Vedclass · Accepted Answer

$12.27$

Vedclass · Answer

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE_{\max})$ of the emitted electron is given by:$KE_{\max} = E - \phi_0$where $E = 5.50 \, eV$ is the photon energy and $\phi_0 = 4.50 \, eV$ is the work function.$KE_{\max} = 5.50 \, eV - 4.50 \, eV = 1.00 \, eV$.The de Broglie wavelength $(\lambda)$ of an electron is given by the formula:$\lambda = \frac{h}{\sqrt{2mE}} = \frac{12.27}{\sqrt{E(eV)}} \, \mathring{A}$.Substituting $E = 1.00 \, eV$:$\lambda = \frac{12.27}{\sqrt{1.00}} \, \mathring{A} = 12.27 \, \mathring{A}$.Since $12.27 \, \mathring{A}$ is not explicitly listed as an option but corresponds to the calculation,and checking the provided options,there appears to be a discrepancy in the provided choices. Based on standard physics,the answer is $12.27 \, \mathring{A}$.

The work function of tungsten is $4.50 \, eV$. The wavelength of the fastest electron emitted when light whose photon energy is $5.50 \, eV$ falls on the tungsten surface is ......... $\mathring{A}$.

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