Chemistry, 29.05.2020 09:57 lilyrockstarmag

In order to comply with the requirement that energy be conserved, Einstein showed in the photoelectric effect that the energy of a photon (h) absorbed by a metal is the sum of the work function (), the minimum energy needed to dislodge an electron from the metal's surface, and the kinetic energy (Ek) of the electron: h = + Ek. When light of wavelength 357.4 nm falls on the surface of potassium metal, the speed (u) of the dislodged electron is 5.8×105 m/s. What is (in kJ/mol) of potassium?

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Answer from: queenkalah101

Complete Question

In order to comply with the requirement that energy be conserved, Einstein showed in the photoelectric effect that the energy of a photon (h) absorbed by a metal is the sum of the work function ( $\phi$ ), the minimum energy needed to dislodge an electron from the metal's surface, and the kinetic energy (Ek) of the electron: $hf = \phi +Ek$ . When light of wavelength 357.4 nm falls on the surface of potassium metal, the speed (u) of the dislodged electron is 5.8×105 m/s. What is work function (in kJ/mol) of potassium?

The work function is $\phi_m = 242.195 \ KJ /mol$

Explanation:

From the question we are told that

$hf = \phi +Ek$

Where h is the Planck's constant which has a constant value of $h = 6.626 *10^{-34} J \cdot s$

Ek is the kinetic energy which is mathematically represented as

$Ek = \frac{1}{2} m v^2$

Where m is the mass of electron with a constant value

$m = 9.11 *10^{-31} \ kg$

The wavelength is $\lambda = 357.4 nm = 357.4 *10^{-9} \ m$

The speed of the electron is $v = 5.8*10^{5} m/s$

This photoelectric effect can be mathematically defined as

f is the frequency of the incident light which is mathematically represented as $f = \frac{c}{\lambda }$

$\phi$ is the work function

So we have

$\frac{hc}{\lambda } = \phi + \frac{1}{2} mv^2$

substituting value

$\frac{6.626 *10^{-34}* 3*10^8}{357.4 *10^{-9}} = \phi + \frac{1}{2} * (9.11*10^{-31}) * (5.8*10^{8}) ^2$

$5.5618*10^{-19} = \phi + 1.54*10^{-14}$

=> $\phi = 4.021*10^{-19} J$

For one mole of one mole the work function is

$\phi _m = \phi * N$

Where N is the avogadro's constant with a value

$N = 6.022*10^{23}$

$\phi_m = 4.021*10^{-19} * 6.022 *10^{23}$

$\phi_m = 242194.90 J /mol$

now in KJ

$\phi_m = 242.195 \ KJ /mol$

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