In the n^{text{th}} Bohr orbit,the ratio of t | vedclass

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(B) In the Bohr model of the atom,the kinetic energy $(K)$ of an electron in the $n^{	ext{th}}$ orbit is given by $K = \frac{kZe^2}{2r_n}$.The total

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$1:-1$

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(B) In the Bohr model of the atom,the kinetic energy $(K)$ of an electron in the $n^{	ext{th}}$ orbit is given by $K = \frac{kZe^2}{2r_n}$.The total energy $(E)$ of the electron in the $n^{	ext{th}}$ orbit is given by $E = -\frac{kZe^2}{2r_n}$.Comparing these two expressions,we find that $K = -E$.Therefore,the ratio of the kinetic energy to the total energy is $\frac{K}{E} = \frac{-E}{E} = -1$.This can be expressed as the ratio $1:-1$.

In the $n^{\text{th}}$ Bohr orbit,the ratio of the kinetic energy of an electron to the total energy of it is:

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