Iron has a work function of 4.50 eV.

(a) What cutoff frequency of light is necessary to cause electrons to be ejected from an iron plate?

At the cutoff frequency we have hf_o = W_o.

This gives:

fo

=

Wo h

=

4.5 eV * 1.6 x 10-19J/eV 6.63 x 10-34 J s

=

1.09 x 1015 Hz

(b) Is this in the visible spectrum?

l

=

c f

=

3 x 108 1.09 x 1015

=

2.76 x 10-7 m

=

276 nm

The smallest wavelength in the visible spectrum is 400 nm. 276 nm is ultraviolet light.

(c) If the iron plate is exposed to light with a frequency of 1.5 x 10¹⁵ Hz, what is the maximum kinetic energy of the ejected electrons?

K_max = E - W_o = hf - W_o

Kmax

=

6.63 x 10-34 J s * 1.5 x 1015 Hz 1.6 x 10-19 J/eV

- Wo

K_max = 6.22 eV - 4.50 eV = 1.72 eV

or K_max = 1.72 eV x 1.6 x 10^-19 J/eV = 2.75 x 10^-19 J

(d) What voltage is necessary to return the electrons to the iron?

1.72 volts.