Consider a beryllium z = 4) ion with all but one of its electrons removed. What is the kinetic energy of the electron in the n = 1 orbit. The radius of the first Bohr orbit in Be³⁺ is 0.0132 nm.) A) 54.4 eV B) 218 eV C) 13.6 eV D) 27.2 eV E) 109 eV

Question

Quizplus · Accepted Answer

The kinetic energy of the electron in the first Bohr orbit can be calculated using the formula: $K.E. = \frac{1}{2}mv^2 = \frac{1}{2}\frac{Ze^2}{4\pi\epsilon_0r}$, where $m$ is the mass of the electron, $v$ is its velocity, $Z$ is the nuclear charge, $e$ is the elementary charge, $\epsilon_0$ is the permittivity of free space, and $r$ is the radius of the first Bohr orbit.

Substituting the given values, we get:

$K.E. = \frac{1}{2}\frac{(4)(1.602	imes10^{-19}	ext{ C})^2}{4\pi(8.854	imes10^{-12}	ext{ C}^2/	ext{N}\cdot	ext{m}^2)(0.0132	imes10^{-9}	ext{ m})}$

$K.E. = 218	ext{ eV}$

Therefore, the correct answer is B.

Consider a Beryllium Z = 4) Ion with All but One