Photoelectric effect and wavelengths

[punjabi_monster]

Here is a question I am having trouble with , thanks for your help.

Electrons are ejected from a photoelectric surface with a maximum speed of 4.20 * 10^5 m/s. If the work function of this surface is 2.55 eV, what is the wavelength of the incident light?

This is how i tried to solve this question:
Ek=(1/2)mv^2
Ek=(1/2)(9.11*10^-31 kg)(4.20*10^5 m/s)^2
Ek=8.04*10^-20 J

E=hc/λ
λ=[(6.63*10^-34 Js)(3.00*10^8 m/s)]/(8.04*10^-20 J)
λ=(2.48*10^-6 m)

The actual answer is (4.07*10^-7 m).

Can you please tell me the correct way to do this question. Thanks.

 

[apchemstudent]

Here is a question I am having trouble with , thanks for your help.

Electrons are ejected from a photoelectric surface with a maximum speed of 4.20 * 10^5 m/s. If the work function of this surface is 2.55 eV, what is the wavelength of the incident light?

Ek=(1/2)mv^2
Ek=(1/2)(9.11*10^-31 kg)(4.20*10^5 m/s)^2
Ek=8.04*10^-20 J

E=hc/λ
λ=[(6.63*10^-34 Js)(3.00*10^8 m/s)]/(8.04*10^-20 J)
λ=(2.48*10^-6 m)

The actual answer is (4.07*10^-7 m).

Can you please tell me the correct way to do this question. Thanks.

Your Ek was correct, but you forgot to include the energy needed to eject an electron from the surface 2.55 eV.

 

[punjabi_monster]

Would you use this formula:
Ekmax = hf - W

 

[quasar987]

That's right.

$$E_{k_{max}}=E_{photons}-W$$

Because the incident photons give away all of their energy to the electrons. So all the electrons have to do to break free from the metal is to escape its "potential well" of energy 2.55 eV.

 

[punjabi_monster]

oooo thanks...i got the right answer!