Chapter 38 Workbook Solutions

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12. The figure shows a stopping potential-versusfrequency
graph for a photoelectric effect experiment.
On the figure, draw and label graphs for the following
two situations:
i. The light intensity is increased,
ii. The cathode work function is increased.
In each case, no other parameters of the experiment
are changed.
13. In a photoelectric effect experiment, the frequency of the light is increased. As a result:
i. There are more photoelectrons. iii. Both i and ii,
Cn.
photoelectrons are faster.
Explain your choice.
E u\n. = hf witl olso
cl,\So irncfeo,.SLS.
iv. Neither i nor ii.
Quantization . cHAPTER 38 38-5
increa,sc o'!e so K^o*= Es1o.- E o
14. ln a photoelectric effect experiment, the intensity of the light is increased. As a result:
Qfn.r.
are more photoelectrons. iii. Both i and ii.
ii. The photoelectrons are faster. iv, Neither i nor ii.
il il:::'n ,^i"^.e ti q\r* d"riuors a, larler nr,rmbcr e{ r,#
euqrnta lo *hc- Si"fo.o ejoclinq q - \orgcr n\ wtbcf oT
;i J. e\ cctr o v,s q.v\ t {horc by co,*,.sin3 q \ 6f* cr.trrenf .
15. A gold cathode (work function = 5.1 eV) is illuminated with light of wavelength 250 nm. It is found
that the photoelectron current is zero when LV = 0 V. Would the current change if:
a. The intensity is doubled?
No. $= "/t = 3xlos'./s,/"so:,lg-i* : l. !.* lgts H e
$o= +. fr$qq = l.!.3* to,S He f .fo
b. The anode-cathode potential difference is increased to LV = 5,5 V?
No. {' fo
c. The cathode is changed to aluminum (work function = 4.3 eV)?
Je-s. + = e/x = 3xlo8'/t/rSo xl0-{",1 =
'r .3 e\
{J\" I o''sNs :1.0'l x\015
l, ).x lols He
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