Question

A photon with a wavelength of 2.47 nm strikes a surface and emits an electron with kinetic energy of 228 eV (electron volts). What is the binding energy of the electron in J? Round your answer to 2 sig figs. 1 electron volt = 1.602 x 10-19 J

Answer #1

The kinetic energy (K.E) of the emitted electron will be equal to the energy of the photon - the binding energy of the electron:.

1/2 mv^{2} = h - E

Given, = 2.47 nm =
2.47 x 10^{-9} m

We know, c =

=> = c/
=( 3 x 10^{8} m/s / 2.47 x 10^{-9} m)

=> = 1.215 x
10^{17} s^{-1}

and 1/2 mv^{2} = 228 ev = 228 x 1.602 x 10^{-19}
J

=> 1/2 mv^{2} = 365.256 x 10^{-19} J = 3.65 x
10^{-17} J

Now, E = h - 1/2
mv^{2}

=> E = (6.626 x 10^{-34} Js x 1.215 x 10^{17}
s^{-1}) -( 3.65 x 10^{-17} J )

=> E = (8.05 x 10^{-17} J) - (3.65 x 10^{-17}
J)

=> E = (8.05 - 3.65) x 10^{-17} J

=> E = 4.4 x 10^{-17} J

Now, The Avogadro's constant , L = 6.022 x 10^{23}
mol^{-1}

So, E = 4.4 x 10^{-17} J x 6.022 x 10^{23}
mol^{-1}

=> E = 26.4968 x 10^{6} J/mol

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Use conservation of (relativistic) energy and momentum to
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