Question

example 4 - 4
the minimum energy needed to remove an electron from a potassium metal surface is 3.7×10^(-19) j. will photons of frequencies 4.3×10^(14) s^(-1) (red light) and of 7.5×10^(14) s^(-1) (blue light) trigger the photoelectric effect? if so, what is the maximum kinetic energy of the ejected electrons?

Explanation:

Step1: Recall photon - energy formula

The energy of a photon is given by $E = h
u$, where $h = 6.63\times10^{-34}\ J\cdot s$ is Planck's constant and $
u$ is the frequency of the photon.

Step2: Calculate energy of red - light photon

For red light with $
u_{red}=4.3\times 10^{14}\ s^{-1}$, $E_{red}=h
u_{red}=6.63\times 10^{-34}\ J\cdot s\times4.3\times 10^{14}\ s^{-1}=2.85\times 10^{-19}\ J$.

Step3: Compare red - light photon energy with binding energy

The binding energy $E_{b}=3.7\times 10^{-19}\ J$. Since $E_{red}=2.85\times 10^{-19}\ J

Step4: Calculate energy of blue - light photon

For blue light with $
u_{blue}=7.5\times 10^{14}\ s^{-1}$, $E_{blue}=h
u_{blue}=6.63\times 10^{-34}\ J\cdot s\times7.5\times 10^{14}\ s^{-1}=4.97\times 10^{-19}\ J$.

Step5: Compare blue - light photon energy with binding energy

Since $E_{blue}=4.97\times 10^{-19}\ J>E_{b}$, blue - light photons will trigger the photoelectric effect.

Step6: Calculate maximum kinetic energy of ejected electrons

According to the photoelectric - effect equation $K_{max}=E_{photon}-E_{b}$. For blue light, $K_{max}=E_{blue}-E_{b}=4.97\times 10^{-19}\ J - 3.7\times 10^{-19}\ J=1.27\times 10^{-19}\ J$.

Answer:

Red - light photons will not trigger the photoelectric effect. Blue - light photons will trigger the photoelectric effect, and the maximum kinetic energy of the ejected electrons is $1.27\times 10^{-19}\ J$.