## Saplings #9

Frankie Mele 3J
Posts: 63
Joined: Wed Sep 30, 2020 10:09 pm

### Saplings #9

Suppose the typical work function of the metal is roughly 5.010×10−19 J. Calculate the maximum wavelength in angstroms of the radiation that will eject electrons from the metal.
Can someone let me know how they solved for the wavelength on this question. I keep getting 3.968 x10^-7 but I'm not sure if I'm maybe converting to angstroms wrong or if the equation I'm using is wrong. Thank you!

Mahnoor_Wani_1I
Posts: 81
Joined: Wed Sep 30, 2020 9:35 pm
Been upvoted: 2 times

### Re: Saplings #9

Here is my thought process (My value given for work function was different than yours):
Given: the work function aka the threshold energy (3.200 x10^-19 J)
We need to solve for max wavelength of E=hv function. Wavelength is inversely proportional to frequency. So max wavelength = min frequency. Min frequency means the energy per photon will be the min energy needed to ejected the electron (aka the threshold energy).

Using the two equations: E= hv and c= wavelength * v
we get wavelength = (h)(c)/ E since the two equation at top both have V

Then we plug in our the parameters and variables and we get the wavelength is roughly 6.21 * 10^-7 meters
To convert it into nm we multiply 6.21 * 10^-7 meters by 1 angstrom/10^-10 m
We get: 6206 Angstrom

I think you probably converted to angstrom wrong. Remember 1 angstrom is 10^-10 meters

Kamille Kibria 2A
Posts: 55
Joined: Wed Sep 30, 2020 9:52 pm

### Re: Saplings #9

uppose the typical work function of the metal is roughly 5.010×10−19 J. Calculate the maximum wavelength in angstroms of the radiation that will eject electrons from the metal.
Can someone let me know how they solved for the wavelength on this question. I keep getting 3.968 x10^-7 but I'm not sure if I'm maybe converting to angstroms wrong or if the equation I'm using is wrong. Thank you!

to solve this i first used the equation (work function=hv0) where v0 is the threshold frequency and h is Planck's constant. you divide the work function by Planck's constant to get the threshold frequency. (5.010x10^-19 J/6.626x10^-34) = 7.561x10^14 Hz. then you use the equation c=lamda(v) to solve for the wavelength. (3x10^8/7.56x10^14) = 3.97x10^-7 m. but angstroms are in units of 10^-10 m, so the answer in angstroms is 3968 angstroms. :)

Frankie Mele 3J
Posts: 63
Joined: Wed Sep 30, 2020 10:09 pm

### Re: Saplings #9

Thank you this helped a lot! I was definitely converting my units wrong.