Photoelectric effect 2

The maximum kinetic energy of an ejected electron is given by $K_{max} = h\left(f - f_0\right)$ , where $h = 6.6261 \times 10^{-34}$ Js is the Planck constant, $f$ Hz is the frequency of the incident photon and $f_0$ Hz is the threshold frequency. Therefore,

$\begin{aligned} K_{max} & = h\left(f - f_0 \right) \\ \Rightarrow f_0 & = f - \frac{K_{max}}{h} \quad \quad \small \color{#3D99F6}{\text{As } \lambda f = c = 2.9979 \times 10^8 \space ms^{-1} \text{, speed of light}} \\ & = \frac{c}{\lambda} - \frac{K_{max}}{h} \\ & = \frac{2.9979 \times 10^8}{400.0 \times 10^{-9}} - \frac{1.54 \times 10^{-19}}{6.6261 \times 10^{-34}} \\ & = 5.1706 \times 10^{14} \space Hz \end{aligned}$

Therefore, the longest wavelength of light $\lambda_0 = \dfrac{f_0}{c} = \dfrac{5.1706 \times 10^{14}}{2.9979 \times 10^8} = \boxed{579.8} \space nm$

The photoelectric effect formula:

$hf = hf_0 + \dfrac{1}{2}m_e(v_{max})^2$

• $h=6.626 \times 10^-34 \text{ J/s}$

• $f$ is the frequency of the incident light

• $f_0$ is the threshold frequency (dependent on the kind of metal)

• $\dfrac{1}{2}m_e(v_{max})^2$ is the maximum kinetic energy of the ejected electrons (electron mass $m_e = 9.11 \times 10^-31 \text{ kg}$ )

The first part of the photoelectric effect involves the incident light, or the photons hitting the metal surface. Those photons are traveling at the speed of light $c = 3 \times 10^8 \text{ m/s}^2$ with a wavelength $\lambda = 400 \text{ nm}$ .

*Note: $400 \text{ nm} = 4 \times 10^{-7} \text{ m}$

$\Rightarrow \textbf {STEP 1:} \text{ Find} f$

We can solve for $f$ , the frequency of the incident light, using $c=f \lambda$ :

$f = \dfrac{c}{\lambda} = \dfrac{3 \times 10^8 \text{ m/s}^2}{4 \times 10^{-7} \text{ m}} = 7.5 \times 10^{14} \text{ Hz}$

$\Rightarrow \textbf {STEP 2:} \text{ Find } v_{max} \text { of ejected electrons}$

We're given that $K_{max} = 1.54 \times 10^{-19} \text{ J}$ , and we know that $m_e = 9.11 \times 10^-31 \text{ kg}$ .

So, using basic algebra, we'll find that $v_{max}=5.81 \times 10^5 \text{ m/s}^2$ .

$\Rightarrow \textbf {STEP 3:} \text{ Solve for} f_0$

Rearranging the very first equation to solve for $f_0$ gives:

$hf_0 = hf - \dfrac{1}{2}m_e(v_{max})^2$

Using the values we found in steps 1 and 2, we can solve the above equation to find that: $f_0=5.176 \times 10^{14} \text{ Hz}$ .

*Note: $\text{ Hz} = \text{ s}^{-1}$

$\Rightarrow \textbf {STEP 4:} \text{ Find } \lambda_0$

The reason $f_0$ is important is because this value is directly related to the kind of metal involved, meaning it will lead us to the longest wavelength that can eject electrons from that same metal .

We know that the speed of light $c$ is unchanging, and we know $f_0$ , so now we just need to solve for the wavelength (we'll call this wavelength $\lambda_0$ ).

$\lambda_0 = \dfrac{3 \times 10^8 \text{ m/s}^2 } {5.176 \times 10^{14} \text{ s}^{-1}} \approx 5.79 \times 10^{-7} \text{ m} \rightarrow \boxed{579 \text{ nm}}$