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 lamda = 2 pi r`. Number of waves made by an electron in Bohr.s orbit is equal to the number of the orbit.
Also, `2 pi r, =2xx3.14xx0.529xx10^(-8)xx4=13.28xx10^(-8) cm`
Since, `n lamda =2 pi tau therefore lamda =(13.28xx10^(-8))/(2)=6.64xx10^(-8) cm =6.64xx10^(-10) m`
`therefore v=(h)/(lamda m)=(6.626xx10^(-34))/(6.64xx10^(-10)xx9.108xx10^(-31))=1.09xx10^(6) ms^(-1)`