2. (10 points) When electromagnetic waves shine on a platinum surface electrons

Question

2. (10 points) When electromagnetic waves shine on a platinum surface electrons are not ejected unless the wavelength of the EM radiation is less than 196 nm. This observation is now interpreted as one photon from the EM radiation hitting one electron in the platinum atom with enough energy to eject the electron from the atom. (a) If the EM radiation has a wavelength greater than 196 nm, is the energy of each photon greater than, less than, or the same as EM radiation with wavelength less than 196 nm? (b) Why are no electrons ejected if the EM radiation has wavelength greater than 196 nm? (c) Write an equation using the conservation of energy to show what happens when a photon incident on a metal surface ejects an electron. (d) What is the work function for platinum in eV? (Hint: At the threshold wavelength or frequency, any ejected electrons have no kinetic energy since all the energy of the photon is used simply to eject the electron from the metal.) (e) When EM radiation with a wavelength of 141 nm shines on the surface, what is the maximum speed of the ejected electrons? (Hint: find the maximum kinetic energy of the ejected electrons to get the maximum speed.) (f) What is the maximum energy of the electrons in part (e) in eV? What would be stopping potential for these electrons? (8) What is the energy of the photons in part (e) in eV? (h) When a more intense beam of EM radiation with a wavelength of 141 nm shines on the surface what will happen to the number of electrons emitted and their maximum energy? (i) If the wavelength of the 141 nm EM radiation is cut in half, what is the new maximum kinetic energy of the electrons in eV? Why cant you solve this question using the same method of ratios that we often use in class and on exams? G) If the wavelength of the 141 nm EM radiation is doubled, what is the new maximunm kinetic energy of the electrons in eV

Explanation / Answer

(A) Energy = h c / wavelength

if wavelength is greater than energy will be less.

Ans: less than

(B) energy at 196 nm is the threshold energy photon required to eject an electron.

if wavelength is greater than energy of photon will be less than threshold energy hence no electron is ejected.

(C) energy of photon = threshold energy to eject electron + kinetic energy of ejected electron

{ h c / wavelength = work function + KE_electron }

(D) work function = energy at 196 nm

= h c / wavelength

= (1240 eV-nm) / 196 nm

= 6.33 eV

(E) E_photon = (1240 / 141) = 8.79 eV

8.79 = 6.33 + KE_e

KE_e = 2.46 eV

(9.11 x 10^-33)v^2 /2 = (2.46 x 10^-19 x 1.6)

v = 9.30 x 10^5 m/s

(F) KE_max = 2.46 eV , stopping potential = 2.46 Volt

(g) E_photon = 8.79 eV

(h) number of electron will increase.

maximum energy will be same.

(i) E_photon = 2(8.79) = 17.6 eV

KE_max = 17.6 - 6.33 = 11.3 eV

(j ) E_photon < work function

hence no electron is ejected

so KE = 0

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