Wavelengths belonging to Balmer series lying in the range of 450 nm to 750 nm were used to eject photoelectrons from a metal surface whose work function is 2 . 0 eV . Find (in eV ) the maximum kinetic energy of the emitted photoelectrons. (Take h c = 1242 eV nm .)

Step-by-step solution

Wavelengths corresponding to minimum wavelength (λ min ) or maximum energy will emit photoeiectrons having kinetic energy which is maximum. (λmin ) belonging to Balmer series and Lying in the given range (450 nm to 750 nm) corresponds to transition from (n = 4 to n = 2). Here, E ⁡ 4 = - 1 3 · 6 4 2 = - 0 · 8 5 eV and E ⁡ 2 = - 1 3 · 6 2 2 = - 3 · 4 eV ∴ Δ E ⁡ = E ⁡ 4 - E ⁡ 2 = 2 · 5 5 eV K ⁡ max = Energy of photon - work function = 2 · 5 5 - 2 · 0 = 0 · 5 5 eV