Correct answer option is (C)

$\frac{1}{2}|PE|=KE$ for each value of $\mathrm{n}$ (orbit)

$\therefore \frac{KE}{|PE|}=\frac{1}{2}$

Correct answer option is (B)

The radius of the nth stationary orbit in the Bohr model of an atom is directly proportional to the square of its principal quantum number n and inversely proportional to the atomic number Z. This relationship is represented by the formula $r_n=\frac{n^2{h}^2}{4{\pi }^{2}km{e}^{2}Z},$ where :

• $n$ is the principal quantum number,

• $h$ is Planck's constant,

• $k$ is the Coulomb constant,

• $m$ is the mass of the electron,

• $e$ is the charge of the electron, and

• $Z$ is the atomic number of the nucleus (for hydrogen, Z=1).

To find the radius of the fourth orbit ($r_4$), in comparison to the third orbit ($r_3$), we apply the formula with $n=4$ for the fourth orbit and $n=3$ for the third orbit, and simplify as follows:

$\begin{array}{rl}& \frac{r_4}{r_3}=\frac{(4^2)h^2}{4{\pi }^{2}km{e}^{2}Z}÷\frac{(3^2)h^2}{4{\pi }^{2}km{e}^{2}Z}=\frac{(4^2)}{(3^2)}\\ \\ & =\frac{16}{9}\end{array}$

Thus, the radius of the fourth stationary orbit is $\frac{16}{9}$ times the radius of the third stationary orbit, $R$. Therefore, $r_4=\frac{16}{9}R$.

Correct answer is 16

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 $\stackrel{o}{A}$), and
• $n$ is the principal quantum number (orbit number).

Given that:

• The radius of a certain orbit of the hydrogen atom is $8.48\stackrel{o}{A}$, 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:

$8.48=n^2\times 0.529$

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,

$x=16$