Radius of a certain orbit of hydrogen atom is 8.48 Å. If energy of electron in this orbit is E/x, then x = ______.
(Given a0 = 0.529Å, E = energy of electron in ground state)
(Given a0 = 0.529Å, E = energy of electron in ground state)
Correct Answer: 16
Approach Solution - 1
We know:
\[ r = 0.529 \frac{n^2}{Z} \implies 8.48 = 0.529 \frac{n^2}{1} \]
\[ n^2 = 16 \implies n = 4 \]
We also know:
\[ E \propto \frac{1}{n^2} \]
\[ E_n = \frac{E}{16} \]
Thus, $x = 16$.
Approach Solution -2
The problem provides the radius of a specific electron orbit in a hydrogen atom and asks for a value 'x' related to the energy of the electron in that same orbit. The energy is given in the form E/x, where E is the ground state energy.
Concept Used:
The solution is based on the Bohr model for the hydrogen atom. According to this model, the radius of the n-th stationary orbit and the energy of the electron in that orbit are quantized and given by the following formulas:
1. Radius of the n-th orbit (\(r_n\)):
\[ r_n = a_0 \frac{n^2}{Z} \]where \(a_0\) is the Bohr radius (approximately 0.529 Å), \(n\) is the principal quantum number, and \(Z\) is the atomic number. For a hydrogen atom, \(Z=1\).
2. Energy of the n-th orbit (\(E_n\)):
\[ E_n = E_1 \frac{Z^2}{n^2} \]where \(E_1\) is the energy of the ground state (\(n=1\)). For a hydrogen atom (\(Z=1\)), and using \(E\) to denote the ground state energy as per the problem, the formula simplifies to:
\[ E_n = \frac{E}{n^2} \]Step-by-Step Solution:
Step 1: Identify the given values and the atomic number for hydrogen.
Given radius of the orbit, \(r_n = 8.48 \text{ Å}\).
Bohr radius, \(a_0 = 0.529 \text{ Å}\).
For the hydrogen atom, the atomic number is \(Z = 1\).
Step 2: Use the formula for the radius of an orbit to determine the principal quantum number, \(n\).
\[ r_n = a_0 \frac{n^2}{Z} \]Substituting the given values and \(Z=1\):
\[ 8.48 \text{ Å} = (0.529 \text{ Å}) \frac{n^2}{1} \]Step 3: Solve for \(n^2\).
\[ n^2 = \frac{8.48}{0.529} \] \[ n^2 = 16 \]Step 4: Find the value of the principal quantum number \(n\).
\[ n = \sqrt{16} = 4 \]So, the electron is in the 4th orbit.
Step 5: Use the formula for the energy of an electron in the n-th orbit to find the energy in this specific orbit (\(E_4\)).
\[ E_n = \frac{E}{n^2} \]Substitute \(n=4\):
\[ E_4 = \frac{E}{4^2} = \frac{E}{16} \]Final Computation & Result:
The problem states that the energy of the electron in this orbit is \(E/x\). We have calculated the energy to be \(E/16\).
By comparing the given expression with our calculated result:
\[ \frac{E}{x} = \frac{E}{16} \]This implies that the value of \(x\) must be 16.
Therefore, the value of x is 16.
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