Question 311: Atoms & Nuclei

Question

Radius of a certain orbit of hydrogen atom is 8.48 $$\mathop A\limits^o$$. If energy of electron in this orbit is $$E / x$$. then $$x=$$ ________ (Given $$\mathrm{a}_0=0.529$$ $$\mathop A\limits^o$$, $$E=$$ energy of electron in ground state).

Accepted Answer

Let's approach this problem by understanding the basics and applying the Bohr model to find the energy levels of a hydrogen atom.
The energy of an electron in a hydrogen atom for any given orbit can be defined using the formula:
$$E_n = \frac{E}{n^2}$$
where:
$E$ is the energy of the electron in the ground state ($n=1$), and
$n$ is the principal quantum number (orbit number).
The radius of an orbit in the hydrogen atom, according to the Bohr model, is given by:
$$r_n = n^2 a_0$$
where:
$r_n$ is the radius of the nth orbit,
$a_0$ is the Bohr radius ($0.529$ angstroms or $\mathop A\limits^o$), and
$n$ is the principal quantum number (orbit number).
Given that:
The radius of a certain orbit of the hydrogen atom is $8.48 \mathop A\limits^o$, and
The energy of an electron in this orbit is $E/x$,
We need to find $x$.
First, let's find $n$, the principal quantum number for the orbit with radius $8.48$ angstroms:
$$8.48 = n^2 	imes 0.529$$
Solving for $n^2$:
$$n^2 = \frac{8.48}{0.529}$$
$$n^2 \approx 16.03$$
For simplicity and practicality in the quantum model, $n^2$ approximately equal to 16 would imply $n = 4$, considering $n$ must be a whole number and $16.03$ is close to $16$, which is a perfect square of $4$.
Now, to find $x$, we'll use the energy relationship. Since the energy levels of the hydrogen atom are inversely proportional to the square of the principal quantum number $n$,
$$E_{orbit} = \frac{E}{n^2} = \frac{E}{4^2} = \frac{E}{16}$$
According to the given information, $E_{orbit} = \frac{E}{x}$, which means:
$$\frac{E}{x} = \frac{E}{16}$$
Hence,
$$x = 16$$

Atoms & Nuclei

Radius of a certain orbit of hydrogen atom is 8.48 $\mathop A\limits^o. If energy of electron in this orbit is E / x. then x= ________ (Given \mathrm{a}_0=0.529 \mathop A\limits^o, E=$ energy of electron in ground state).

Integer / Numeric Answer Type