1-Light with a frequency of 7.24 x 1014 Hz strikes the surface of a piece of met
ID: 2015593 • Letter: 1
Question
1-Light with a frequency of 7.24 x 1014 Hz strikes the surface of a piece of metallic sodium.What is the energy (in eV) of each photon of this light beam?
2-Light with a frequency of 6.71 x 1014 Hz strikes the surface of a piece of metallic sodium.
For sodium, the energy needed to tear an electron out of the metal surface, or work function, is 2.46 eV.
What is the maximum kinetic energy (in eV) that an electron may have after being knocked out of the surface by a single photon in the beam described above?
3-An electron coasts into a region where an electric field is present and eventually comes to rest. If the electric potential changed by 2.8 V from the entry point to the stopping point, what was the original kinetic energy (in eV) of the electron?
4-What is the longest wavelengths (in nanometers) that we can use to knock electrons out of sodium?
Explanation / Answer
1- Frequency of light is (f) = 7.24x1014 Hz from Einsein energyequation energy E = h f where h is plank's constant is = 6.625 x 10-34 J.S required energy is E= 6.625 x 10-34x7.24x1014 = 47.965 x 10-20J 2- Given frequency f =6.71 x 1014 Hz work funtion (w) =2.46 eV = 2.46 x 1.6 x 10-19J = 3.936x10-19J where 1eV = 1.6x10-19J from photo electric effect conservation of energy principle energy E = maximum kinetic energy + work funtion (w)..........(1) but energy E = h f =6.625 x 10-34x6.71 x 1014 E = 44.45375 x10-20J...............(2) from equation (1) and (2) 44.45375 x10-20J = K.E +3.936 x 10-19J Required kinetic energy K.E =0.51775x10-19J =0.3235 eV 3- Given that potential difference (V) =2.8V Required kinetic energy is given by K.E = e xV where e is charge of electron is given by e = 1.6x10-19 C Required kinetic energy is K.E = 1.6x10-19 x2.8 =4.48 x10-19J =4.48 eV = 47.965 x 10-20J 2- Given frequency f =6.71 x 1014 Hz work funtion (w) =2.46 eV = 2.46 x 1.6 x 10-19J = 3.936x10-19J where 1eV = 1.6x10-19J from photo electric effect conservation of energy principle energy E = maximum kinetic energy + work funtion (w)..........(1) but energy E = h f =6.625 x 10-34x6.71 x 1014 E = 44.45375 x10-20J...............(2) from equation (1) and (2) 44.45375 x10-20J = K.E +3.936 x 10-19J Required kinetic energy K.E =0.51775x10-19J =0.3235 eV 3- Given that potential difference (V) =2.8V Required kinetic energy is given by K.E = e xV where e is charge of electron is given by e = 1.6x10-19 C Required kinetic energy is K.E = 1.6x10-19 x2.8 =4.48 x10-19J =4.48 eVRelated Questions
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