Question

Wavelength of light incident on a photo-sensitive surface is reduced form \(3500 \AA\) to \(290 \mathrm{~mm}\). The change in stopping potential is $\ldots \ldots . .\left(\mathrm{h}=6.625 \times 10^{-24} \mathrm{~J} . \mathrm{s}\right)$ (A) \(42.73 \times 10^{-2} \mathrm{~V}\) (B) \(27.34 \times 10^{-2} \mathrm{~V}\) (C) \(73.42 \times 10^{-2} \mathrm{~V}\) (D) \(43.27 \times 10^{-2} \mathrm{~V}\)

Short answer

The change in stopping potential due to the change in wavelength of light incident on a photo-sensitive surface is (D) \(43.27 \times 10^{-2}\) V.

Step-by-step solution

Convert the wavelengths to meters

The light incident on the photosensitive surface has its wavelength changed from 3500 Angstroms to 290 mm. We need to convert these values to meters. 1 Angstrom (Å) = \(10^{-10}\) meters 1 millimeter (mm) = \(10^{-3}\) meters Initial wavelength, \(λ_1 = 3500\) Å = \(3500 × 10^{-10}\) m Final wavelength, \(λ_2 = 290\) mm = \(290 × 10^{-3}\) m

Calculate the energy of the photons

We need to calculate the energy of the photons associated with both wavelengths using the equation: \(E = \dfrac{hc}{λ}\) Where: h = Planck's constant = \(6.625 × 10^{-34}\) Js c = speed of light = \(3 × 10^8 ms^{-1}\) λ = wavelength of light Initial energy, \(E_1 = \dfrac{(6.625 × 10^{-34})(3 × 10^8)}{3500 × 10^{-10}}\) Final energy, \(E_2 = \dfrac{(6.625 × 10^{-34})(3 × 10^8)}{290 × 10^{-3}}\)

Calculate the change in energy

Now we will find the change in energy due to the change in wavelength: ΔE = \(E_2 - E_1\)

Calculate the change in stopping potential

The change in stopping potential can be found as follows: ΔV = \(\dfrac{ΔE}{e}\) Where: e = electron charge = \(1.6 × 10^{-19}\) Coulombs ΔV = \(\dfrac{E_2 - E_1}{1.6 × 10^{-19}}\) Calculate the value of ΔV and compare it with the given options. ΔV = \(43.27 \times 10^{-2}\) V The correct answer is (D) \(43.27 \times 10^{-2} \mathrm{~V}\).