a metallic surface is illuminated with light of wavelength 400 nm. if the work function for this metal is 2.4 eV. what is the maximum kinetic energy of the ejected electrons, in electron volts?

Assume that
1 eV = 1.6 * 10^-19
c = 3 * 10^8
h = 6.626 * 10^-34

I already got the answer before and I know that the max kinetic energy is 0.71 eV but when I tried to solve it again, I forgot what to do.

I tried getting the cut off wave lengths and using that combined with the wavelength provided to get the K.E max but it didn't work

The kinetic energy (KE) of electrons is defined as the product of one-half of the mass of the electron to the square of the velocity at which electrons spin in orbit.

Given data;

λ(Wavelength)= 400 × 10⁻⁹ m

Φ(Work function)=2.4 eV

c(Speed of light)= 3 ×10⁸ m/sec

h(Constant) = 6.626 * 10^-34

[tex]\rm \phi = 2.4 eV = 2.4 \times 1.6 \times 10^{-19} \\\\ \rm \phi =3.84 \times 10^{-19} \ J[/tex]

The maximum kinetic energy is found as;

[tex]\rm KE= \frac{hc}{\lambda} - \phi \\\\ KE= \frac{6.64 \times 10^{-34}}{400 \times 10^{-9}} -3.84 \times 10^{-19} \\\\ KE=1.14\times 10^{-19} \ J[/tex]

Unit conversion:

1 eV = 1.6 * 10⁻¹⁹ j

1 J=1/( 1.6 * 10⁻¹⁹) ev

KE=0.71 eV

Hence, the maximum kinetic energy of the ejected electrons will be 0.71 eV.

To learn more about the kinetic energy of electrons refers to:

https://brainly.com/question/13145345

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