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////Given
lembda=3500*10**-10 //m
h=6.6*10**-34
c=3*10**8 //m/s
//calculation
E=((h*c)/lembda)/1.6*10**-19
//Result
printf("\n Energy is %0.2f ev",E*10**38)
printf("\n 1.9 ev < E < 4.2 ev,only metal B will yield photoelectrons")
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