QJEEM24Phe0e6: JEE Main 2024 Physics - Class 12 JEEM24Phe0e6 Solution: Radius of a certain orbit of hydrogen atom is 8.48 . If energy…

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:

$8.48 = n^2 \times 0.529$

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,

$x = 16$