Answer to Question #290326 in Mechanics | Relativity for Arnica

Question

A sodium surface is illuminated with light of wavelength 0.300 mm. The work function for sodium is 2.46 eV. Calculate

(a) the energy of each photon in electron volts,

(b) the maximum kinetic energy of the ejected photoelectrons, and

(c) the cutoff wavelength for sodium.

Answer 1

(a) Calculate the energy of each photon.

Obtain the frequency from the wavelength:

$\begin{aligned}&c=f \lambda \quad \rightarrow \quad f=\frac{c}{\lambda}=\frac{3.00 \times 10^{8} m / s }{0.300 \times 10^{-6} m } \\&f=1.00 \times 10^{15} Hz\end{aligned} \\ E ​ =hf=(6.63×10 ^{−34} J⋅s)(1.00×10^{ 15} Hz) \\ =6.63×10^{ −19} J \\ =(6.63×10^{ −19} J)( \frac{1.00eV}{1.60×10^{ −19} J} ​ )=4.14 \; eV ​$

(b) Find the maximum kinetic energy of the photoelectrons.

$KE max ​ =hf−ϕ=4.14eV−2.46eV=1.68eV$

(c) Compute the cutoff wavelength.

Convert $\phi$ from electron volts to joules:

$ϕ ​ =2.46eV=(2.46eV)(1.60×10^{ −19} J/eV) \\ =3.94×10^{ −19} J ​$

​Find the cutoff wavelength

$λ c ​ ​ = \frac{hc}{ ϕ} \\ ​ = \frac{(6.63×10^{ −34} J⋅s)(3.00×10^ 8 m/s)}{3.94×10^{ −19} J} \\ ​ =5.05×10^{ −7} m=505\; nm ​$