The wavelength of an electron moving in the second orbit of H-atom is an integral multiple of its circumference. Calculate
speed of electron in the second orbit

Explanation

Circumference is 2`pi`. Wavelength is an intergral multiple of the circumference. Thus, `n lambda=2pi r`. Number of waves made by an electron in Bohr.s orbit is equal to the number of the orbit.
a) Number of waves (n) in the second orbit=2
b) Also `2pi r_(2)=2xx3.14xx0.529xx10^(-8)xx4=13.28xx10^(-8)cm`
Since, `n lambda=2pi r :.lambda=(13.28xx10^(-8))/(2)=6.64xx10^(-8)cm =6.64 xx 10^(-10)m`
`:.v=(h)/(lambda m)=(6.626xx10^(-34))/(6.64xx10^(-10)xx 9.108xx10^(-31))=1.09xx10^(6)ms^(-1)`
c) Also `eV_(0)=(1)/(2)mv^(2) " " :. V_(0)=(9.108xx10^(-31)xx(1.09xx10^(6))^(2))/(2xx1.602xx10^(-19))=3.38V`