Question

The stopping potential in a photoelectric experiment is linearly related to the inverse of the wavelength (1/λ) of the light falling on the cathode. The potential difference applied across an X-ray tube is linearly related to the inverse of the cutoff wavelength (1/λ) of the X-ray emitted. Show that the slopes of the lines in the two cases are equal and find its value.

Answer 1

V₀ - Stopping Potential
K - Potential difference across X-ray tube
λ - Wavelength
λ - Cut difference Wavelength
$$\begin{gathered} e{\mathrm{V}}_0=hf-h{f}_0 \\ \lambda =\frac{hc}{e\mathrm{V}} \\ e{\mathrm{V}}_0=\frac{hc}{\lambda } \\ \mathrm{or}\mathrm{V\lambda }=\frac{hc}{e} \\ \mathrm{or}{\mathrm{V}}_0\mathrm{\lambda }=\frac{hc}{e} \end{gathered}$$
Here, the slopes are same.
i.e. V₀λ = Vλ
$$\begin{gathered} \frac{hc}{e}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}} \\ =1.242\times {10}^{-6}\mathrm{Vm} \end{gathered}$$