In the Bohr's model of hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in `n^(th)` quantum state is:

Options

A. `-1`
B. `+1`
C. `-2`
D. `+2`

Answer

A

Explanation

In Bohr.s model of hydrogen atom,
The kinteic energy of the electron in `n^("th")` state is given by
`K=(me^(4))/(8epsilon_(0)^(2)h^(2)n^(2))=(13.6)/(n^(2))eV` when `(me^(4))/(8epsilon_(0)^(2)h^(2))=13.6eV`
The potential energy of electron in `n^("th")` state is given by
`U=(-2me^(4))/(8epsilon_(0)^(2)h^(2)n^(2))=(--272)/(n^(2))eV`
Total energy of electron in `n^("th")` state is given by
`E=K+U=(me^(4))/(8epsilon_(0)^(2)h^(2)n^(2))-(2me^(4))/(8epsilon_(0)^(2)h^(2)n^(2))=(-me^(4))/(8epsilon_(0)^(2)h^(2)n^(2))=(-13.6)/(n^(2))eV`
`therefore (K)//(E)=-1`