which is equivalent to the result shown in the problem statement. Since the desired numerical answer uses eV units, we multiply numerator and denominator of our result by c2and make use of the mc2value for an electron in Table 37-3 as well as the value 1240eV nmhc=⋅: NEmchcnnnF()(((.)///=FHGIKJ=×⋅FHGIKJ−−434 511 101240341122231332321ππeV)eV nm)nmeV2which is equivalent to the value indicated in the problem statement. (b) Since there are 1027cubic nanometers in a cubic meter, then the result of Problem 41-5 may be written as n=×=849 108492833...mnmThe cube root of this is n1/3≈4.4/nm. Hence, the expression in part (a) leads to
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