Question #120364

VII. The photoelectric threshold wavelength of a tungsten surface is 272 nm. Calculate the maximum

kinetic energy of the electrons ejected from this tungsten surface by ultraviolet radiation of frequency

1.45 x 1015 Hz. Express the answer in electron volts.

Notes:

Remember that E=hf=hc/λ

Threshold wavelength means the minimum amount of wavelength to remove electrons from a surface

so in this problem, E = = hf ϕ threshold=hc/ λthreshold

where λthreshold = threshold wavelength

= work function = the minimum energy needed to remove an electron from the surface ϕ

h = Planck’s constant = 6.626 x 10 -34 J.s = 4.136 x 10 -15 eV.s

c = 3x108 m/s

f = frequency.

kinetic energy of the electrons ejected from this tungsten surface by ultraviolet radiation of frequency

1.45 x 1015 Hz. Express the answer in electron volts.

Notes:

Remember that E=hf=hc/λ

Threshold wavelength means the minimum amount of wavelength to remove electrons from a surface

so in this problem, E = = hf ϕ threshold=hc/ λthreshold

where λthreshold = threshold wavelength

= work function = the minimum energy needed to remove an electron from the surface ϕ

h = Planck’s constant = 6.626 x 10 -34 J.s = 4.136 x 10 -15 eV.s

c = 3x108 m/s

f = frequency.

Expert's answer

"\\frac{hc}{\\lambda}=A+KE"

"A=\\frac{hc}{\\lambda_0}=\\frac{6.62\\cdot 10^{-34}\\cdot 3\\cdot10^8}{272\\cdot10^{-9}}=7.3\\cdot10^{-19}J"

"KE=h\\nu-7.3\\cdot10^{-19}=6.62\\cdot 10^{-34}\\cdot 1.45\\cdot10^{15}-7.3\\cdot10^{-19}="

"=2.3\\cdot10^{-19}J=1.44eV"

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