A student studying the photoelectric effect from twodifferent metals records the
ID: 1727256 • Letter: A
Question
A student studying the photoelectric effect from twodifferent metals records the following information: (i) thestopping potential for photoelectrons released from metal 1 is 1.48V larger than that for metal 2, and (ii) the threshold frequencyfor metal 1 is 40.0% smaller than that for metal 2. Determine thework function for each metal. What would be the final answer? Details: Given stopping potential for metal-1 = 1.48 V stopping potentialfor metal - 2 = V Thresholdfreqency for metal -1 = - 0.4 = 0.6 Thresholdfrequency of metal -2 = There fore the relation between stoppingpotential , threshold frequency and work function eV = h f - substitute the values inthe above equation so , For metal - 1 e ( V + 1.48 ) = h c / 0.6 -1 The work function of metal -1 1 = hc / 0.6 - e (V+1.48 ) For metal - 2 e V = h c / - 2 The work function of metal -2 2 = h c / - eV where h = plancksconstant = 6.63 x 10-34 Js c = speed of light = 3 x 108 m/s e = charge of electron = 1.6 x 10-19 C A student studying the photoelectric effect from twodifferent metals records the following information: (i) thestopping potential for photoelectrons released from metal 1 is 1.48V larger than that for metal 2, and (ii) the threshold frequencyfor metal 1 is 40.0% smaller than that for metal 2. Determine thework function for each metal. What would be the final answer? Details: Given stopping potential for metal-1 = 1.48 V stopping potentialfor metal - 2 = V Thresholdfreqency for metal -1 = - 0.4 = 0.6 Thresholdfrequency of metal -2 = There fore the relation between stoppingpotential , threshold frequency and work function eV = h f - substitute the values inthe above equation so , For metal - 1 e ( V + 1.48 ) = h c / 0.6 -1 The work function of metal -1 1 = hc / 0.6 - e (V+1.48 ) For metal - 2 e V = h c / - 2 The work function of metal -2 2 = h c / - eV where h = plancksconstant = 6.63 x 10-34 Js c = speed of light = 3 x 108 m/s e = charge of electron = 1.6 x 10-19 C Given stopping potential for metal-1 = 1.48 V stopping potentialfor metal - 2 = V Thresholdfreqency for metal -1 = - 0.4 = 0.6 Thresholdfrequency of metal -2 = There fore the relation between stoppingpotential , threshold frequency and work function eV = h f - substitute the values inthe above equation so , For metal - 1 e ( V + 1.48 ) = h c / 0.6 -1 The work function of metal -1 1 = hc / 0.6 - e (V+1.48 ) For metal - 2 e V = h c / - 2 The work function of metal -2 2 = h c / - eV where h = plancksconstant = 6.63 x 10-34 Js c = speed of light = 3 x 108 m/s e = charge of electron = 1.6 x 10-19 CExplanation / Answer
hf - W1 = e1 hf - W2 = e21 -2 = 1.48 V 1.48 eV = W2 - W1 hfs1 = W1 hfs2 = W2fs1 =0.6fs2 W1 = 0.6W2 1.48 eV = 0.4W2 W2 = 3.7 eV W1 = 2.22 eV
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