In a photoelectric experiment, the stopping potential exhibits a...

From Einstein's photoelectric equation, the stopping potential $V_0$ is given by:

$eV_0 = \dfrac{hc}{\lambda} - \Phi$

$V_0 = \dfrac{hc}{e} \left(\dfrac{1}{\lambda}\right) - \dfrac{\Phi}{e}$

The slope of the graph of $V_0$ versus $1/\lambda$ is $\dfrac{hc}{e}$.

For an X-ray tube, the cutoff wavelength $\lambda_c$ is given by:

$eV = \dfrac{hc}{\lambda_c}$

$V = \dfrac{hc}{e}$\left(\dfrac{1}{\lambda_c}\right)

The slope of the graph of $V$ versus $1/\lambda_c$ is also \dfrac{hc}{e}.

The common slope is \dfrac{hc}{e}.

Substituting the standard values $h = 6.626 \times 10^{-34} \text{ J s}$, $c = 3 \times 10^8 \text{ m/s}$, and $e = 1.6 \times 10^{-19} \text{ C}$:

$\text{Slope} =$\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19}}

$\text{Slope} = 12.42 \times 10^{-7} \text{ V m} = 1.242 \times 10^{-6} \text{ V m}$