Question

Calculate the wavelength of light, in nm, illuminating a barium surface if the threshold frequency of barium is 6.07x10^14 HZ and the kinetic energy of the electron emitted is 1.31x10^-18.

Answer 1

@timo86m @karatechopper

Answer 2

\[\phi _{Barium}=E _{light}-E _{electron}\]
\[\phi _{Barium}=hf _{Barium}\]
\[E _{light}=hf _{light}\]

Answer 3

\[hf _{Ba}=hf _{Light}-E _{electron} \]

Answer 4

\[c=\lambda f \rightarrow f=c/\lambda\]
\[hf _{Ba}=hc/\lambda_{light}-E _{electron}\]
solve for lambda, the wavelength in meters:
\[\lambda _{light}= \frac{ hc}{hf _{Ba}+E _{electron}} \]
h = 6.63*10^-34 J*s
c = 3.00*10^8 m/s
fba = 6.07*10^14 1/s
Eelectron=1.31x10^-18 J

Plug in, solve, and convert to nm by multiplying by (10^9 nm / 1 m)

Answer should be 116 nm

Answer 5

You are a life saver!

Answer 6

Thank you very much.

Answer 7

np it was a good refresher