Answer to Question #233107 in Atomic and Nuclear Physics for Ebuski

Question

What is the longest-wavelength EM radiation that can eject a photoelectron from silver, given that the binding energy is 4.73 eV?

Answer 1

The kinetic energy of an ejected electron is given as

$KE_e=hf -BE$

Here,

h is the Plank’s constant,

f is the frequency of material andis the binding energy of the electron.

As speed of light is equal to the product of frequency of light and wavelength of light then it is given as,

$c=fλ$

Here,

f is the frequency of light and

λ is the wavelength of light.

Rearrange the above equation for f as follows:

$c=fλ \\ f=\frac{c}{λ }$

$KE_e=hf -BE \\ KE_e= h(\frac{c}{λ }) -BE$

Rearrange the equation in terms of wavelength λ.

$KE_e+BE= \frac{hc}{λ } \\ λ = \frac{hc}{KE_e+BE}$

Here, the maximum kinetic energy of an electron ejected using the light of threshold frequency ($KE_e$ ) is zero. Which means the out of all the electrons ejected using the light of threshold frequency the one with the highest kinetic energy is zero.

Substitute:

$h=4.14 \times 10^{-15} \;eV \cdot s \\ c = 3.00 \times 10^8 \;m/s \\ KE_e=0 \\ BE = 4.73 \;eV \\ λ = \frac{(4.14 \times 10^{-15} \;eV \cdot s)(3.00 \times 10^8 \;m/s)}{(0+4.73 \;eV)} \\ = 2.63 \times 10^{-7} \; m \\ = 263 \;nm$

Hence, the magnitude of longest wavelength is 263 nm.

No, the wavelength is not in the visible range. This is in ultraviolet range.