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Determine the de Broglie wavelength of a proton, whose kinetice energy is equal to the rest of the mass energy of an electron. Given that the mass of an electron is `9xx10^(-31)` kg and the mass of a proton is `1837` times as that of the electron. |
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Answer» Here mass of electron, `m_0=9.1xx10^(-31)`kg Mass of proton, `m=1837xx9.1xx10^(-31)=1.67xx10^(-27)`kg let v be the velocity of the proton. Then, `(1)/(2)mv^(2)=m_0c^(2)` or `m^(2)v^(2)=2mm_0c^(2)` or `mv=sqrt(2mm_0c)` `=sqrt(2xx1.67xx10^(-27)xx9.1xx10^(31))xx3xx10^(8)` `=1.654xx10^(20)kgms^(-1)` Threfore, de Broglie wavelength of the proton, `lamda=(h)/(mv)=(6.62xx10^(-32))/(1.654xx10^(-20))=4xx10^(-14)m` |
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