Question

Photons with a frequency of 8.9 1015 Hz shine on a piece of platinum (work function = 6.4 eV). What is the energy of the ejected electrons?

What is the de Broglie wavelength of the ejected electrons?

Answer #1

**energy of incident photon E = h*f = 6.626*10^-34*8.9*10^15
= 5.89*10^-18 J**

**work function Wo = 6.4 eV = 6.4*1.6*10^-19 =
1.024*10^-18 J**

**from photo electric equation**

**E = Wo + KE**

**energy of emitted electrons KE = E - Wo = 5.89*10^-18 -
1.024*10^-18 = 4.866*10^-18 J**

**==============================**

**KE , K= (1/2)*m*^2**

**speed v = sqrt(2*K*m)**

**De brogile wavelength lambda = h/mv =
h/sqrt(2*K*m)**

**lambda = 6.626*10^-34/(sqrt(2*4.866*10^-8*9.11*10^-31) =
2.22*10^-15 m**

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