A photon with 250.0 keV encounters a neutral atom of^ 197Au. The work function o

Question

A photon with 250.0 keV encounters a neutral atom of^ 197Au. The work function of Au is A = 5.1 eV, which is the binding energy of an electron in the outermost shell. (a) How many electrons does the neutral ^197 Au atom have? (b) What is the maximum energy of a photoelectron emitted from this nuclide? (c) If the photon Compton scatters from an electron in the outer shell through an angle of 45 degrees, what will the scattered photon's energy be? /30 points] (d) What is the recoil kinetic energy of the scattered electron? [30 points] E= hv- A

Explanation / Answer

(a) 197Au nucleide has atomic number 79. For any neutral species (atom) number of electrons are equal to its atomic number hence the neytral atom of 197Au has 79 electrons.

(b) For emitted photon,

Energy of incident photon = 250 KeV = 250 x 103 eV and A= binding energy of outermost shel electron = 5.1 eV.

Maximum energy of emitted photon = Energy of incident photon - Binding energy of electron in outermostshell

Hence, Max. energy of emitted photn = 250 x 103 - 5.1 = 2.49994 x 105

(b) For emited photon,

Energy of incoming photon (hv) = 250 keV = 250 x 103 eV...... (small v pronounce as nu).

Angle of scattering T = 45o and cos T = cos 45o = 0.707

Energy of emmited photoelectron(hv') = ? (say E)

Formula,

E = {hv x Ax (1-cosT) } / [1+ A(1-cosT).....................(1)

with CosT= 0.707 hv = 250 KeV we get,

E = 250 x A x (1-0.707) /[1+A(1-0.707).

E = (250 x A x 0.293) / (1+0.293A)........................(2)

Here A (alpha) = hv/mco2 = 250 / 0.511 x 103   ... (mco2 = 0.511 MeV= 0.511 x 103 KeV

So. A = 0.4892 put it into eq.2

E = 250x 0.4892 x 0.293 / (1+0.4892 x 0.293)

E = 35.8339 / 1.1433

E = 31.3435 KeV.

Scattered photon energy = 31.3435 KeV.

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