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## Homework Statement

When light of wavelength 298 nm is incident on potassium, the emitted electrons have

maximum kinetic energy of 1.65 eV.

A)What is the energy of an incident photon?

The value of h c is 1240 eV · nm .

Answer in units of eV.

B)What is the work function for potassium?

Answer in units of eV.

C)

What would be the maximum kinetic energy

of the electrons if the incident light had a

wavelength of 375 nm?

Answer in units of eV.

D)

What is the threshold wavelength for the photoelectric eﬀect with potassium?

Answer in units of nm.

## Homework Equations

E=H*V

E=H*F

## The Attempt at a Solution

A)

this is the part that is confusing, it gives us the HC which should mean that i have the energy of the incident photon according to law of conservation of energy. but surely it can't be that simple, this is an extra chapter we didn't get to so i have to go by what i find online. according to http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html it is a totally different number

any assistance would be appreciated