Question

the critical wavelength for producing the photoelectric effect in tungsten is 2600 angstrom. what wavelength would be necessary to produce the photoelectrons from tungsten having twice the KE of these produced at 2200 angstrom?

Answer 1

So you have \[KE_2 = 2KE_1\], is what you'll need, notice it talks about the photoelectric effect, so we're dealing with light, \[E_{photon} = hf = \frac{ hc }{ \lambda }\], you should now be able to put it all together, good luck!