Textbook Problem 1.31

[Kaya Lu 2K]

I was trying to complete #31 of the cumulative problems for Focus 1, but I am stuck on how to solve part A of this problem. I checked the solution manual but I am very confused about where they got the J*eV^-1 value they multiply the work function by to obtain the energy needed to eject an electron from the lithium. Does anyone know how to proceed in solving this problem?

[emmarolland2B]

Hi! Here is how I did this problem.
First, I converted the given work function (2.93 eV) into J by using the conversion facter 1 eV = 1.602 x 10^-19 J.
2.93 eV x (1.602 x 10^-19J/1 eV) = 4.69 x 10^-19 J.

Next, we want to determine the energies of both lasers because we need to determine which laser has an energy greater than or equal to the work function, as this is the amount of energy needed to eject the electron.

To find the energy of each photon, I used E=hc/wavelength

For red ruby laser:
1) convert the wavelength into m
694 nm = 694 x 10^-9m
2) plug values into the equation:
E=hc/wavelength
E=(6.626x10^-34)(3.0x10^8)/(694x10^-9)
E= 2.86 x 10^-19 J

For the violet laser:
1) convert the wavelength into m
405 nm = 405 x 10^-9 m
2) plug values into the equation:
E=hc/wavelength
E=(6.626x10^-34)(3.0x10^8)/(405x10^-9)
E=4.9 x 10^-19 J

In this case, the violet laser has an energy that is greater than the work function (4.69 x 10^-19 J), while the red ruby laser has less energy than the work function. Therefore, the violet laser is better to use in this case because it has enough energy to eject the electron from the metal. I hope this helps!