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प्रश्न
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 × 108 m/s.)
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उत्तर
V0 - 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. V0λ = Vλ
`(hc)/e = (6.63 xx 10^-34 xx 3 xx 10^8)/(1.6 xx 10^-19)`
= `1.242 xx 10^-6 "Vm"`
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