In a photoelectric effect experiment, the stopping potential is...

According to Einstein's photoelectric equation:

$V_s = \dfrac{hc}{e\lambda} - \Phi$

Using the data points $\lambda_1 = 350 \text{ nm}$, $V_{s1} = 1.45 \text{ V}$ and $\lambda_2 = 400 \text{ nm}$, $V_{s2} = 1.00 \text{ V}$:

$1.45 = \dfrac{hc}{350e} - \Phi$

$1.00 = \dfrac{hc}{400e} - \Phi$

Subtracting the second equation from the first:

0.45 = \dfrac{hc}{e} \left(\dfrac{1}{350} - \dfrac{1}{400} \right)

$0.45 =$\dfrac{hc}{e} \left( \dfrac{400 - 350}{350 \times 400} \right) = \dfrac{hc}{e} \left( \dfrac{50}{140000} \right)$<br><br>$\dfrac{hc}{e} = 0.45 \times 2800 = 1260 \text{ V nm}$<br><br>Substituting this value back into the second equation to find the work function$\Phi$:<br><br>$1.00 = \dfrac{1260}{400} - \Phi$<br><br>$\Phi = 3.15 - 1.00 = 2.15 \text{ eV}$<br><br>The threshold wavelength$\lambda_0$is given by:<br><br>$\lambda_0 = \dfrac{hc}{e\Phi}$<br><br>$\lambda_0 = \dfrac{1260}{2.15} \approx 586 \text{ nm}$<br><br>This value is closest to$585 \text{ nm}$.<br><br>Answer:$585 \text{ nm}