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The wavelength of light from the spectral emission line of sodium is 589nm. Find the kinetic energy at which (a) an electron, and (b) a neutron, would have the same de-broglie wavelength. |
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Answer» Solution :Here `lamda=589nm=5.89xx10^(-7)m` From the relation `lamda=(h)/(sqrt(mK))`, we have `K=(h^(2))/(2mlamda^(2))` (a) For an electron `m=9.11xx10^(-31)KM` `THEREFORE K_(c)=((6.63xx10^(-34))^(2))/(2xx1.675xx10^(-27)xx(5.89xx10^(-7))^(2))=3.78xx10^(-28)J=(3.78xx10^(-28))/(1.6xx10^(-19))eV=2.36xx10^(-9)eV` `=2.36` neV. |
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