Light of wavelength 200 nm falls on an aluminum surfaceIn aluminum 4.2 eV are required to remove an electronWhat is the kinetic energy of (a) the fastest and (b) the lowest emitted photoelectrons? (c) What is the stopping potential? (d) What is the cutoff wavelength for aluminum?
Hover just above your chair inside a squat position for 15 mere seconds every hour.
these can be found from your chapter on photo electric effect energy of photon will liberate the electron it will lose 4.2 eV in moving to the surface than it will move with kinetic energy of whatever the energy is left so the photon is 200 nm E h f f c/L then E h c/L h 4.14110^-15 eV units E 4.14110^-15310^8/20010^-9 E 2.0705310^0 E 6.2115 eV E K + Wo K E - Wo K 6.2- 4.2 2.0 eV so the highest K will be 2.0 eV the lowest K will be of course 0 eV (because it could hit more than the usual number of atoms and loose all its energyc) what is the stopping potential if the highest Ke of the electron is 2.0 eV then to stop any electrons from having any Ke we just need to have a force pushing them back with 2.0 eV potential I would think this is -2 V W eV solve for V this means if the V is reversed it will have enough Electric field to stop the electrons with the highest Ke from hitting the anode V K/e V 2eV/e 2V so a stopping voltage of -2V is required to stop the electrons it must be negative because electrons are repelled by a negatively charged plate d) the cutoff wavelength is just the lowest energy photon that will just excite the electron E Wo since hc/L Wo solve for L L hc/Wo use h 4.14110^-15 L 2.95810^-7 L 295.8 nm thats all there is to it just think it through
