UA practice problem frequency of light

Moderators: Chem_Mod, Chem_Admin

Linda Arroyo 3J
Posts: 15
Joined: Fri Apr 06, 2018 11:01 am

UA practice problem frequency of light

Postby Linda Arroyo 3J » Thu Oct 25, 2018 10:01 pm

6. The work function for lithium is 4.6 x 10−19 J.
a. What is the lowest frequency of light that will cause photoelectric emission?
The ANswer is 6.9 x 1014 Hz

I am just confused on what equation they used to solve this.

Xinyi Zeng 4C
Posts: 63
Joined: Fri Sep 28, 2018 12:18 am

Re: UA practice problem frequency of light

Postby Xinyi Zeng 4C » Thu Oct 25, 2018 10:08 pm

First, consider E (photon)=hv, for frequency of light, v to be the lowest, E should be the lowest possible, too.
Now think about E(photon) - work function = E(excess) = Kinetic energy of the ejected electron
If we want E(photon) to be the lowest, we can take the E(excess) to be zero. Then, E (photon) = work function of Lithium.
Then, since we know E(photon), you can just use E=hv to solve for frequency.

Karina Jiayu Xu 4E
Posts: 58
Joined: Fri Sep 28, 2018 12:29 am

Re: UA practice problem frequency of light

Postby Karina Jiayu Xu 4E » Sun Oct 28, 2018 1:21 pm

Can someone explain what a work function is? I think i got that wrong on the test because i thought a work function was work as a function of time.

Linh Vo 2J
Posts: 61
Joined: Sat Apr 28, 2018 3:00 am

Re: UA practice problem frequency of light

Postby Linh Vo 2J » Tue Nov 06, 2018 3:01 am

Karina Jiayu Xu 4E wrote:Can someone explain what a work function is? I think i got that wrong on the test because i thought a work function was work as a function of time.


Yes I can. Work function is basically the idea of the energy required to eject or emit an electron. In the equation, E(photon) = work function + E(kinetic energy). Basically, the work function is the minimum amount of energy needed to eject an electron from a metallic surface (usually an ultraviolet wavelength). Work function is not work as a function of time. Work function can be denoted with a big O with a line striked through it.


Return to “Photoelectric Effect”

Who is online

Users browsing this forum: No registered users and 2 guests