Question

Photons with a frequency of 3e15 Hz shine on a piece of aluminum (work function = 5.4 eV). What is the energy of the ejected electrons?

=____J

What is the de Broglie wavelength of the ejected electrons?

=_____nm

Answer 1

a)let the energy of the ejected electrons be x.so,

so,

Energy = energy of the photons-work function

=6.626*10^-34*3*10^15/(1.6*10^-19) - 5.4

=7.02 eV

=1.12*10^-18 J

b)let the wavelength be x.so,

x=h/(2*m*KE)^0.5

=6.626*10^-34/(2*7.02*1.6*10^-19*9.1*10^-31)^0.5

=4.63*10^-10 m

=4.63 nm