Question

The stopping potential in a photoelectric experiment is linearly related to the inverse of the wavelength $\frac{1}{\lambda }$ 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 $\frac{1}{\lambda }$ of the X-ray emitted. Show that the slopes of the lines in the two cases are equal and find its value.

Answer 1

$V_0$ -Stopping Potential

K - Potential difference across X-ray tube

$\lambda$ - Wavelength

$\lambda$ - Cut difference Wavelength

$e{V}_0=hf-h{f}_0$

$\lambda =\frac{hc}{eV}$

$e{V}_0=\frac{hc}{\lambda }$

or, V $\lambda =\frac{hc}{e}$

or, $V_0\lambda =\frac{hc}{e}$

Here Slopes are same

i.e $V_0\lambda =V\lambda$

$\frac{hc}{e}=\frac{6.63\times {10}^{-34\times 3\times {10}^8}}{1.6\times {10}^{-19}}$

=$1.242\times {10}^{-6}$ Vm