- #1
frankene
- 6
- 0
Light is incident on the surface of metallic sodium, whose work function is 2.3 eV. The maximum speed of the photoelectrons emitted by the surface is 1.26e6 m/s. What is the wavelength of the light?
I first converted Work Function to Joules:
2.3eV x (1.6e-19 J / 1eV) = 3.68e-19 J
The equation I know is:
hf = KEmax + Work Function
hf can also be written as:
hc/lamda
I thought I would substitute c w/ the 1.26e6 m/s and use KEmax as zero. My answer was 2.27e-9m and that is not correct.
I used the following:
f = Work Function / h = 5.55e14 Hz
Then I used:
lamda = v/f = 2.27e-9 m
If I can't substitute v for c, then what am I looking for with v? I have looked at this problem for 3 days w/ no new ideas. Anyone else have any?
Thank you!
I first converted Work Function to Joules:
2.3eV x (1.6e-19 J / 1eV) = 3.68e-19 J
The equation I know is:
hf = KEmax + Work Function
hf can also be written as:
hc/lamda
I thought I would substitute c w/ the 1.26e6 m/s and use KEmax as zero. My answer was 2.27e-9m and that is not correct.
I used the following:
f = Work Function / h = 5.55e14 Hz
Then I used:
lamda = v/f = 2.27e-9 m
If I can't substitute v for c, then what am I looking for with v? I have looked at this problem for 3 days w/ no new ideas. Anyone else have any?
Thank you!