A gold cathode with work function 5.1 eV is illuminated with light of wavelength 250 nmIt is found that the photoelectron current is zero when the anode-cathode potential difference is zeroWould the current change if the1)light intensity is doubled? Explain your answer.2)anode-cathode potential difference is increased to 5.5 V? Justify your answer.3)cathode is changed to aluminum (work function 4.3 eV)? Explain your answer.
1) current increases in generalEmitted electrons have some initial k.eand they move towards the anode to give rise to a currentAs the intensity increases the number of electrons emitted increase and hence the current increasesIn the particular case hf hc/λ 6.625x10^-34x3x10^8x6.624x10^18/250x10^-4.96 eV , which is less than the work function 5.1 eVHence there is no photoelectric emissionSo no current 2) Increases to a saturation value in generalIncrease in the anode cathode potential increases the speed of the electrons or the noof electrons crossing unit area per second, hence the current increasesBeyond a certain value of the accelerating potential the velocity of the electrons becomes maximum and cannot increase further , and the current reaches a saturation valueIn the above case since there is no photoelectric current no question change in current3)The photoelectric equation is hf W + mv^2/2 So in general for a given frequency if the work function W is smaller or decreased then the k.eof the electrons is more and the speed is moreHence the current increasesIn the above problem the work function is changed to 4.3 eV which is less than hf 4.96 eVHence photo electrons are generated and there is a current even in the absence of accelerating potential.
