Question

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.

Answer #1

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.

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View Available Hint(s)
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