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.

(Use Planck constant h = 6.63 × 10^{-34} Js= 4.14 × 10^{-15} eVs, speed of light c = 3 × 10^{8} m/s.)

#### Solution

V_{0} - Stopping Potential

K - Potential difference across X-ray tube

λ - Wavelength

λ - Cut difference Wavelength

`eV_0 = hf - hf_0`

`lambda = (hc)/(eV)`

`eV_0 = (hc)/lambda`

or ` Vlambda = (hc)/e`

or `V_0lambda = (hc)/e`

Here, the slopes are same.

i.e. V_{0}λ = Vλ

`(hc)/e = (6.63 xx 10^-34 xx 3 xx 10^8)/(1.6 xx 10^-19)`

= `1.242 xx 10^-6 "Vm"`