When light with a wavelength of 258 nm is incident on a certain
metal surface, electrons are ejected with a maximum kinetic energy
of 3.02 × 10-19 J. Determine the wavelength (in nm) of
light that should be used to double the maximum kinetic energy of
the electrons ejected from this surface.
the energy of the photon is initially =
E = h*f = 6.62*10^-34Js*3*10^8m/s /258*10^-9m J
E = 7.69*10^-19 J.
the electron's energy was 3.02*10^-19 J,
so the difference = 4.67*10^-19 J is the work to free the
electron.
To double the energy of the electron
2*3.02*10^-19 + 4.67*10^-19 J = 10.71`10^-19 J energy/photon.
Which has a frequency of
10.71*10^-19 J/ 6.62*10^-34 Js
= 1.617*10^15 Hz
and a wavelength of
3*10^8/1.617*10^15
= 185*10^-9 m
= 185 nm <---- ans aproximately.
Get Answers For Free
Most questions answered within 1 hours.