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Grade 11Physical Chemistry

Calculate energy of electron which is moving in the orbit that has its radius sixteen times the radius of first bohr orbit of H-atom

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8 Years agoGrade 11
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ApprovedApproved Tutor Answer1 Year ago

To find the energy of an electron moving in an orbit that has a radius sixteen times that of the first Bohr orbit of a hydrogen atom, we can use the principles of the Bohr model of the atom. The energy levels of an electron in a hydrogen-like atom can be calculated using a specific formula derived from Bohr's theory.

Understanding the Bohr Model

The Bohr model describes the electron's motion in quantized orbits around the nucleus. For hydrogen, the energy of an electron in the nth orbit is given by the formula:

E_n = -\frac{Z^2 \cdot k \cdot e^4}{2 \cdot n^2 \cdot \hbar^2}

Where:

  • E_n: Energy of the electron in the nth orbit
  • Z: Atomic number (for hydrogen, Z = 1)
  • k: Coulomb's constant
  • e: Elementary charge
  • n: Principal quantum number (1 for the first orbit, 2 for the second, etc.)
  • \hbar: Reduced Planck's constant

Finding the Radius of the First Bohr Orbit

The radius of the first Bohr orbit (n=1) for hydrogen is given by:

r_1 = \frac{n^2 \cdot \hbar^2}{Z \cdot k \cdot e^2}

For hydrogen (Z = 1), this simplifies to:

r_1 = \frac{\hbar^2}{k \cdot e^2}

Calculating the Radius of the New Orbit

If the radius of the new orbit is sixteen times the first orbit, we have:

r = 16 \cdot r_1

Determining the Principal Quantum Number

Since the radius of the nth orbit is given by:

r_n = n^2 \cdot r_1

We can set up the equation:

16 \cdot r_1 = n^2 \cdot r_1

Dividing both sides by r_1 gives:

16 = n^2

Thus, n = 4.

Calculating the Energy of the Electron

Now that we know the principal quantum number is 4, we can substitute n into the energy formula:

E_4 = -\frac{Z^2 \cdot k \cdot e^4}{2 \cdot 4^2 \cdot \hbar^2}

For hydrogen (Z = 1), this simplifies to:

E_4 = -\frac{k \cdot e^4}{32 \cdot \hbar^2}

The energy of the electron in the fourth orbit can be numerically evaluated using known constants:

E_4 ≈ -0.85 eV

Final Thoughts

In summary, the energy of an electron in an orbit with a radius sixteen times that of the first Bohr orbit of hydrogen is approximately -0.85 eV. This negative value indicates that the electron is bound to the nucleus, and the more negative the energy, the more tightly it is bound. Understanding these concepts helps us grasp the fundamental nature of atomic structure and electron behavior.