Calculate the energy of photon in eV for a radiation of wavelength 5000 A ∘ . ( h = 6 . 625 × 10 - 34 J s , c = 3 × 10 8 m s - 1 , e = 1 . 6 × 10 - 19 C )

The energy of a photon is given as E = h ν , here h is Planck constant, and ν is the frequency. If the wavelength is λ , then h ν = h c λ . Now, putting the values in the above expression h = 6 . 625 × 10 - 34 J s , c = 3 × 10 8 m s - 1 , e = 1 . 6 × 10 - 19 C , λ = 5000 A ∘ . E = 6 . 625 × 10 - 34 J s × 3 × 10 8 m s - 1 5000 × 10 - 10 m = 3 . 975 × 10 - 22 J Now, 1 J = 6 . 242 × 10 18 eV Therefore, energy in eV will be 3 . 975 × 10 - 22 × 6 . 242 × 10 18 = 24 . 8 × 10 - 4 eV .

Calculate the energy of photon in eV for a radiation of wavelength 5000 A ∘ . ( h = 6 . 625 × 10 - 34 J s , c = 3 × 10 8 m s - 1 , e = 1 . 6 × 10 - 19 C )

The energy of a photon is given as E = h ν , here h is Planck constant, and ν is the frequency. If the wavelength is λ , then h ν = h c λ . Now, putting the values in the above expression h = 6 . 625 × 10 - 34 J s , c = 3 × 10 8 m s - 1 , e = 1 . 6 × 10 - 19 C , λ = 5000 A ∘ . E = 6 . 625 × 10 - 34 J s × 3 × 10 8 m s - 1 5000 × 10 - 10 m = 3 . 975 × 10 - 22 J Now, 1 J = 6 . 242 × 10 18 eV Therefore, energy in eV will be 3 . 975 × 10 - 22 × 6 . 242 × 10 18 = 24 . 8 × 10 - 4 eV .

The wavelength of light incident on a photosensitive surface is reduced from 3500 ° A to 290 nm . Find change in stopping potential. ( h = 6 . 625 × 10 - 34 Js .)

Given the first wavelength λ 1 = 3500 × 10 - 10 m The second wavelength λ 2 = 290 nm = 2900 × 10 - 10 m From Einstein's equation for photoelectric emission, we get, KE max = eV 1 = hc λ 1 - ϕ and eV 2 = hc λ 2 - ϕ ., where V 1 , V 2 are the stopping potentials corresponding to the wavelengths λ 2 and λ 2 Therefore, the change in stooping potential is given by, e V 1 - V 2 = hc 1 λ 2 - 1 λ 2 or V 1 - V 2 = hc e 1 λ 2 - 1 λ 2 Or V 1 - V 2 = hc e λ 1 - λ 2 λ 1 λ 2 . Substituting the values, we get, V 1 - V 2 = 6 . 625 × 10 - 34 1 . 6 × 10 - 19 3500 × 10 - 10 - 2 . 9 × 10 - 10 3500 × 10 - 10 × 2909 × 10 - 10 . = 12 . 47 × 10 - 7 × 10 10 600 1015 × 10 4 = 0 . 733 V . Hence the change in the stopping potential is 73 . 3 × 10 - 2 V .