Question 461: Atoms & Nuclei

Question

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

Accepted Answer

To find the ratio of the kinetic energy (K.E.) to the total energy (E) of an electron in the
$n^{	ext {th }}$
Bohr orbit, let's first recall the expressions for kinetic energy, potential energy, and total energy in a Bohr orbit.
The kinetic energy of an electron in the
$n^{	ext {th }}$
orbit is given by:
$K.E. = \frac{k e^2}{2r_n}$
where
$k$
is Coulomb's constant,
$e$
is the charge of an electron, and
$r_n$
is the radius of the
$n^{	ext {th }}$
orbit.
The potential energy (P.E.) of an electron in that orbit (due to the electrostatic force of attraction between the positively charged nucleus and the negatively charged electron) is:
$P.E. = - \frac{k e^2}{r_n}$
This expression is negative because the potential energy in an electrostatically bound system is considered zero at an infinite distance and negative within the system due to the attraction.
The total energy (E) of an electron in the
$n^{	ext {th }}$
Bohr orbit is the sum of its kinetic energy and potential energy:
$E = K.E. + P.E. = \frac{k e^2}{2r_n} - \frac{k e^2}{r_n}$
$E = - \frac{k e^2}{2r_n}$
Notice that the total energy is
$- \frac{1}{2}$
times the kinetic energy and also half of the potential energy in magnitude but with a negative sign, indicating that the electron is bound to the nucleus.
To find the ratio of kinetic energy to total energy:
$ \frac{K.E.}{E} = \frac{\frac{k e^2}{2r_n}}{-\frac{k e^2}{2r_n}}$
This simplifies to:
$ \frac{K.E.}{E} = -1$
Therefore, the correct answer is Option B), which represents the ratio of the kinetic energy of an electron to the total energy of it in the
$n^{	ext {th }}$
Bohr orbit as
$1: -1$
.

Answer

$$2: 1$$

Answer

$$1:-1$$

Answer

$$+1: 1$$

Answer

$$-1: 2$$

Atoms & Nuclei

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

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