Question

1. a physicist is asked to analyze two mystery metals. each metal is struck by light with three different wavelengths: 300 nm, 400 nm, and 500 nm. in each case, the physicist adjusts the voltage gradient to determine the energies of the emitted electrons. the results are shown in the chart below. what are the values of x and y?

wavelength	mystery metal a	mystery metal b
400 nm	x	0.8 ev
500 nm	1.3 ev	y

○ x = 1.9 ev, y = 0.2 ev
○ x = 2.1 ev, y cannot be determined
○ x = 2.1 ev, y = 0.2 ev
○ x = 1.9 ev, y cannot be determined

Explanation:

Step1: Recall the photoelectric - effect equation

The energy of a photon is given by $E = h
u=\frac{hc}{\lambda}$, where $h = 4.14\times10^{-15}\text{ eV}\cdot\text{s}$, $c = 3\times10^{8}\text{ m/s}$, and $\lambda$ is the wavelength of the light.

Step2: Calculate the energy of photons for each wavelength

For $\lambda_1 = 300\text{ nm}=300\times10^{-9}\text{ m}$, $E_1=\frac{hc}{\lambda_1}=\frac{4.14\times 10^{-15}\text{ eV}\cdot\text{s}\times3\times10^{8}\text{ m/s}}{300\times10^{-9}\text{ m}} = 4.14\text{ eV}$.
For $\lambda_2 = 400\text{ nm}=400\times10^{-9}\text{ m}$, $E_2=\frac{hc}{\lambda_2}=\frac{4.14\times 10^{-15}\text{ eV}\cdot\text{s}\times3\times10^{8}\text{ m/s}}{400\times10^{-9}\text{ m}} = 3.105\text{ eV}$.
For $\lambda_3 = 500\text{ nm}=500\times10^{-9}\text{ m}$, $E_3=\frac{hc}{\lambda_3}=\frac{4.14\times 10^{-15}\text{ eV}\cdot\text{s}\times3\times10^{8}\text{ m/s}}{500\times10^{-9}\text{ m}} = 2.484\text{ eV}$.

Step3: Use the photoelectric - effect formula $K_{max}=h

u-\phi$
For metal A, when $\lambda = 300\text{ nm}$, $K_{max1}=2.9\text{ eV}$, so $\phi_A=E_1 - K_{max1}=4.14\text{ eV}-2.9\text{ eV}=1.24\text{ eV}$.
When $\lambda = 400\text{ nm}$ for metal A, $K_{maxA2}=E_2-\phi_A=3.105\text{ eV}-1.24\text{ eV}=1.865\text{ eV}\approx1.9\text{ eV}$, so $X = 1.9\text{ eV}$.
For metal B, when $\lambda = 300\text{ nm}$, $K_{maxB1}=1.8\text{ eV}$, so $\phi_B=E_1 - K_{maxB1}=4.14\text{ eV}-1.8\text{ eV}=2.34\text{ eV}$.
When $\lambda = 500\text{ nm}$ for metal B, $K_{maxB3}=E_3-\phi_B=2.484\text{ eV}-2.34\text{ eV}=0.144\text{ eV}\approx0.2\text{ eV}$, so $Y = 0.2\text{ eV}$.

Answer:

$X = 1.9\text{ eV}, Y = 0.2\text{ eV}$