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 mv² = h - E

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

We know, c =

=> = c/ =( 3 x 10⁸ m/s / 2.47 x 10^-9 m)

=> = 1.215 x 10¹⁷ s^-1

and 1/2 mv² = 228 ev = 228 x 1.602 x 10^-19 J

=> 1/2 mv² = 365.256 x 10^-19 J = 3.65 x 10^-17 J

Now, E = h - 1/2 mv²

=> E = (6.626 x 10^-34 Js x 1.215 x 10¹⁷ 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 10²³ mol^-1

So, E = 4.4 x 10^-17 J x 6.022 x 10²³ mol^-1

=> E = 26.4968 x 10⁶ J/mol