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.

Explanation

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.