Electrons with a kinetic energy of 6.023 time | vedclass

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Question

(A) The kinetic energy $(KE)$ of $1 \ mol$ of electrons is $6.023 	imes 10^4 \ J$.Therefore,the $KE$ of $1$ electron is $\frac{6.023 	imes 10^4 \ J/

Vedclass · Accepted Answer

$2.313 	imes 10^{-19} \ J$

Vedclass · Answer

(A) The kinetic energy $(KE)$ of $1 \ mol$ of electrons is $6.023 	imes 10^4 \ J$.Therefore,the $KE$ of $1$ electron is $\frac{6.023 	imes 10^4 \ J/mol}{6.023 	imes 10^{23} \ mol^{-1}} = 1.0 	imes 10^{-19} \ J$.The energy of the incident photon $(E)$ is given by $E = \frac{hc}{\lambda}$.$E = \frac{6.626 	imes 10^{-34} \ J \cdot s 	imes 3 	imes 10^8 \ m/s}{600 	imes 10^{-9} \ m} = 3.313 	imes 10^{-19} \ J$.According to the photoelectric effect equation,$E = \Phi + KE$,where $\Phi$ is the work function (threshold energy).$\Phi = E - KE = 3.313 	imes 10^{-19} \ J - 1.0 	imes 10^{-19} \ J = 2.313 	imes 10^{-19} \ J$.

Electrons with a kinetic energy of $6.023 \times 10^4 \ J/mol$ are evolved from the surface of a metal when it is exposed to radiation of wavelength $600 \ nm$. The minimum amount of energy required to remove an electron from the metal atom is

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