Radiation from a hydrogen-discharge tube passes through a filter that transmits only wavelengths greater than 440 nm . This transmitted radiation is then incident on a metal with a work function of 2.0 eV . Determine the stopping potential required to halt the photoelectrons.

Step-by-step solution

The energy of photons emitted by a hydrogen atom is given by: E = 13.6 ( 1 n₁^2 - 1 n₂^2 ) eV The corresponding wavelength is = 1240 eV nm E . Let us check the transitions for the Balmer series ( n₁ = 2 ): For n₂ = 3 n₁ = 2 : E = 13.6 ( 1 4 - 1 9 ) = 1.89 eV = 1240 1.89 656 nm For n₂ = 4 n₁ = 2 : E = 13.6 ( 1 4 - 1 16 ) = 2.55 eV = 1240 2.55 486 nm For n₂ = 5 n₁ = 2 : E = 13.6 ( 1 4 - 1 25 ) = 2.86 eV = 1240 2.86 434 nm The filter transmits only wavelengths > 440 nm . Thus, the n=5 2 transition and all higher energ