Homework Problems 1.33 and 1.37

[Stevie Wisz]

I am working on completing all of the assigned homework problems and have come across some such as numbers 33 and 37 who have been very difficult for me. This is because Lavelle has not gone over these types of calculations. Will we be required to know how to calculate these types of problem on our first quiz next week?

[Rebecca Sine 1K]

For #1.33, we need to use the "de Broglie" equation, which is "lamda=h/(m*v)". Lambda is wavelength, "h" is Planck's Constant (6.626*10^-34 J*s), "m" is the mass of an electron (9.10939*10^-31 kg), and v is the velocity of the electron.

(a) We are given the velocity of the electron, which is 3.6*10^3 km/s, or 3.6*10^6 m/s. So since we have Planck's constant, mass, and velocity, we will be solving for wavelength.

Lambda= (6.626*10^-34 J*s)/((9.10939*10^-31 kg)*(3.6*10^6 m/s)) = 2.0*10^-10m or 0.2nm.

(b) 2.50*10^16 Hz is the minimum energy required to eject an electron from the surface of a certain metal. We need to find the energy associated with this ejection. Therefore we can use the equation E=h/"nu". "Nu" (the curvy-looking v) is frequency and is measured in Hz (or 1/s). We know "h" and we are given the value of the frequency of the electron, and therefore we are solving for energy E.

E= (6.626*10^-34 J*s)/(2.50*10^16 1/s) = 1.66*10^-17 J

(c) Now we need to figure out the wavelength of the photon that caused the electron to eject in the first place. First we will find the energy of that photon. Knowing the relationship between the energy of an ejected electron and the energy of its incident photon, we can use the equation:

Energy of photon= energy of electron + ((1/2)*(mv^2))

Again, m is mass and v is velocity. We already know the energy of the electron and its velocity from previously, and therefore we are able to solve for the energy of the photon.

Energy of photon= (1.66*10^-17 J) + ((1/2)*(9.10939*10^-31 kg)*(3.6*10^6 m/s)^2) = 2.55*10^-17 J

Now that we know the energy of the photon, we can know find its wavelength using the equation:

E= (h*c)/lambda, where “c” is the speed of light.
Rearrange to solve for wavelength: lambda= (h*c)/E
Lambda= ((6.626*10^-34 J*s)*(3.00*10^8 m/s))/(2.55*10^-17 J)
Lambda= 8.8*10^-9 m, or 8.8nm

(d) The chart on the bottom of page 4 of the chemistry textbook can be used to find out which region of the spectrum corresponds to a wavelength of 8.8nm