Physics, asked by akhilavijayan2877, 1 year ago

Energy of an electron in nth orbit of hydrogen atom is  \bigg \lgroup K=\frac{1}{4\pi\epsilon_0} \bigg \rgroup
(a)  -\frac{2\pi^2K^2me^4}{n^2h^2}
(b)  -\frac{4\pi^2m ke^2}{n^2h^2}
(c)  -\frac{n^2h^2}{2\pi k me^4}
(d)  -\frac{n^2h^2}{4\pi^2 k me^2}

Answers

Answered by phadshyam426
0

SbhshshshzhzjHJjabnabsbsbsbanana

Similar questions