Homework Problems

Moderators: Chem_Mod, Chem_Admin

Posts: 30
Joined: Mon Feb 25, 2019 12:15 am

Homework Problems

Postby Danieljm1B » Tue Jul 09, 2019 12:12 am

Can someone help guide me through the problem 1B. 15 in the 7th edition of the textbook? I am very confused on how to solve it

Katherine Fitzgerald 1A
Posts: 40
Joined: Fri Sep 29, 2017 7:05 am

Re: Homework Problems

Postby Katherine Fitzgerald 1A » Tue Jul 09, 2019 1:26 pm

(A) first identify that you need to use the De Broglie equation to find the wavelength, since an electron is a particle with mass. The wavelength of the electron = h / ( mass of electron * velocity of the electron). Plug in 6.626 * 10^-34 for h, 9.1095 * 10^31 kg for mass of electron, and the given velocity. Make sure you convert the velocity to meters per second do that your units cancel at the end! Solve for wavelength and you should get 2.6 x 10^9

(B) The frequency given is for the threshold energy. Use E=h*frequency to find the energy of the incoming light. This is 6.626*10^-34 joules*sec * 2.5*10^16*(1/s) = 1.656x10^-17

(C) first, find the kinetic energy of the ejected electron (1/2 mass of electron * velocity ^2). Add this to the energy required to remove the electron, which is your answer from C. This is now the total energy of the incoming light that ejected the electron with the velocity given in the problem statement. Set this value of E equal to h*frequency and solve for frequency. Then use this frequency to solve for wavelength is king the relation c=frequency*wavelength

(D) to answer this question, you must know the energies or wavelengths of the different types of light. Compare your answer from C to the energies or wavelengths of light to find a match!


Posts: 50
Joined: Mon Jun 17, 2019 7:22 am

Re: Homework Problems

Postby simmoneokamoto3K » Sun Jul 14, 2019 4:56 pm

First use the De Broglie equation as appropriate turning (lambda) = h/p into (lambda)= h/mv. Next convert the mass of an electron using your conversions, so here are 1000 g in one kg. Once converted you should be left with 9.10939x10^-31. Afterwards convert km into meters so that you're also using the correct units. Once this is done plug all the values into h/mv.

Because the equation asks for energy too in order to solve for the final product we use he equation E = hv as well however as velocity is already a given we just need to plug in our values once more, giving us 1.66 x 10^7 J.

In order to eject the electron we know we need to move it at 3.6 x 10^3 km/s which is our velocity. So in other words we're using 1/2mv^2.

By plugging in all the values the equation shoudl look something like 1/2(mass)(velocity)^2. However we also add 1.66 x 10^-17 becasue we want to know the total energy in order to eject an electron giving us, 2.25x10^-17 J. We have our energy now, however we need the wavelength so once again by combining two equations, E = hv and c = v(lambda) we get E = hc/(lambda) and plug in all of our values giving us 8.8 x 10^-9 = (lambda) or 8.8 nm

Return to “Photoelectric Effect”

Who is online

Users browsing this forum: No registered users and 3 guests