Question #233107

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



1
Expert's answer
2021-09-04T07:35:18-0400

The kinetic energy of an ejected electron is given as

KEe=hfBEKE_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=fλc=fλ

Here,

f is the frequency of light and

λ is the wavelength of light.

Rearrange the above equation for f as follows:

c=fλf=cλc=fλ \\ f=\frac{c}{λ }

KEe=hfBEKEe=h(cλ)BEKE_e=hf -BE \\ KE_e= h(\frac{c}{λ }) -BE

Rearrange the equation in terms of wavelength λ.

KEe+BE=hcλλ=hcKEe+BEKE_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 (KEeKE_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=4.14×1015  eVsc=3.00×108  m/sKEe=0BE=4.73  eVλ=(4.14×1015  eVs)(3.00×108  m/s)(0+4.73  eV)=2.63×107  m=263  nmh=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.


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