A basic question about Photoelectric effect

[docnet]

Homework Statement

Light of wavelength 500 nm is incident on sodium, with work function 2.28 eV. What is the maximum kinetic energy of the ejected photo- electrons?

Relevant Equations

$KE=\frac{hc}{\lambda}-Work$

The energy of a photon with wavelength $\lambda=500nm$ is given by the equation

$E_{photon}=h\nu=\frac{hc}{\lambda}$.
$E_{photon}=3.614\times10^-19 J=2.256eV$

The kinetic energy of an ejected photo-electron is given by

$KE_{electron}=E_{photon}-Work$

Using the given $\lambda$ and the kinetic energy of the photon,
$KE_{electron}=2.256eV-2.28eV=-.024eV$

The maximum kinetic energy of the ejected photo-electron is $-.024eV$
The energy of the photon is insufficient to eject an electron from sodium.

 

[haruspex]

I looked up 500nm at https://www.omnicalculator.com/physics/photon-energy and got 2.48eV.

 

[docnet]

I looked up 500nm at https://www.omnicalculator.com/physics/photon-energy and got 2.48eV.

Thank you! I am getting pretty careless, because I think I put in $550nm$ into the calculator instead of $500nm$, which explains the discrepancy.
$\frac {(6.626\times10^{-34} \frac{m^2kg}{s})(3\times10^8 \frac{m}{s})}{(5\times10^{-7}m)(1.602\times10^{-19} \frac{eV}{J})}=2.48eV$

The correct solution is
$KE_{electron}=2.48eV-2.28eV=.20eV$
The maximum kinetic energy of the ejected photo-electron is $.20eV$.

 

[mjc123]

When you got a crazy answer, didn't you do the calculation again to check?

 

[haruspex]

When you got a crazy answer, didn't you do the calculation again to check?

No emission is not a crazy answer. It could have been a trick question.
But yes, certainly should trigger a careful recomputation or two. Or cheat, like I did.