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The de Broglie wavelength of an electron in a metal at `27^(@)C` is `("Given"m_(e)=9.1xx10^(-31)kg,k_(B)=1.38xx10^(-23)JK^(-1))`A. `6.2xx10^(-9)m`B. `6.2xx10^(-10)m`C. `6.2xx10^(-8)m`D. `6.2xx10^(-7)m` |
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Answer» Correct Answer - A (a) : Here, T=27+273=300K For an electron in a metal, momentum `p=sqrt(3mk_(B)T)` de Broglie wavelength of an electron is `lamda=(h)/(p)=sqrt(3mk_(B)T)` `=(h)/sqrt(3xx(9.1xx10^(-31))xx(1.38xx10^(-23))xx300)` `=6.2xx10^(-9)m` |
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