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An electron in hydrogen atom revolving in a radius of 5.29 xx 10^(-11) m around the nucleus. According to the condition of Bohr's allowed electron orbits, find the principle quantum number corresponds to this orbit. h=6.625xx10^(-34)Js, e=1.6xx10^(-19)C. epsi_(0)=8.85xx10^(-12)MKS, m=9.1xx10^(-31)kg Find out conclusion from your answer. |
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Answer» Solution :For hydrogen atom Z = 1 and centripetal force PROVIDED by coulomb force to electron, `(mv^(2))/(r)=(1)/(4pi epsi_(0))(e^(2))/(r^(2))( :.Z=1)` `:.v^(2)=(e^(2))/(4pi epsi_(0) mr)` Here LINEAR momentum `p=mv :. P= sqrt((me^(2))/(4pi epsi_(0)r))` de-Broglie wavelength, `lambda=(h)/(p)=(h)/(e) sqrt((4pi epsi_(0)r)/(m))` `=(6.625xx10^(-34))/(1.6xx10^(-19))[(4xx3.14xx8.85xx10^(-12)xx5.29xx10^(-11))/(9.1xx10^(-31))]^((1)/(2))` `=4.141xx10^(-15)xx[64.617xx10^(8)]^((1)/(2))` `=33.28xx10^(-11)` `=lambda=3.328x10^(-10)m` `=3.328A^(@)` A According to the condition for Bohr.s allowed electro orbits, `2pi r= n lambda` `:. n=(2pi r)/(lambda)=(2xx3.14xx5.29xx10^(-11))/(3.328xx10^(-10))` `=9.98xx10^(-1)` `~~1` Hence, according to Bohr.s quantum condition the electron is in it ground state (n = 1) while it can be said from de-Broglie hypothesis that ONE de-Broglie wavelength fits for a given RANGE of circumference. |
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