Atoms

1. Question:

Derive the expression for the radius of the nth orbit in the Bohr model of the hydrogen atom.

Solution: According to Bohr's model, the centripetal force on an electron in an orbit is provided by the electrostatic force of attraction between the electron and the nucleus.

$$\frac{mv^2}{r} = \frac{ke^2}{r^2}$$

Where:

• $m$ is the mass of the electron,
• $v$ is the velocity of the electron,
• $r$ is the radius of the orbit,
• $k$ is Coulomb's constant, and
• $e$ is the charge of the electron.

Solving for $v^2$ :

$$mv^2 = \frac{ke^2}{r}$$

From Bohr's quantization rule, the angular momentum of the electron is quantized:

$$mvr = \frac{nh}{2\pi}$$

Where $n$ is the principal quantum number and $h$ is Planck's constant.

Solving for $v$ :

$$v = \frac{nh}{2\pi m r}$$

Substituting this into the centripetal force equation:

$$\frac{m\left(\frac{nh}{2\pi m r}\right)^2}{r} = \frac{ke^2}{r^2}$$

Simplifying:

$$\frac{n^2 h^2}{4\pi^2 m r^3} = \frac{ke^2}{r^2}$$

Solving for $r$ :

$$r = \frac{4\pi^2 m h^2}{k e^2 n^2}$$

This is the expression for the radius of the nth orbit.

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2. Question:

Calculate the radius of the first orbit of the hydrogen atom.

Solution: Using the formula derived above:

$$r = \frac{4\pi^2 m h^2}{k e^2 n^2}$$

For the first orbit ($n = 1$ ):

• $m = 9.11 \times 10^{-31} \, \text{kg}$ ,
• $h = 6.626 \times 10^{-34} \, \text{J·s}$ ,
• $k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2$ ,
• $e = 1.6 \times 10^{-19} \, \text{C}$ ,
• $n = 1$ .

Substituting the values:

$$r = \frac{4\pi^2 (9.11 \times 10^{-31}) (6.626 \times 10^{-34})^2}{9 \times 10^9 (1.6 \times 10^{-19})^2}$$

After simplification:

$$r = 5.29 \times 10^{-11} \, \text{m}$$

Hence, the radius of the first orbit of the hydrogen atom is $5.29 \times 10^{-11} \, \text{m}$ .

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3. Question:

Derive the expression for the energy of the nth orbit in the Bohr model.

Solution: The total energy of the electron in the nth orbit is the sum of its kinetic energy and potential energy.

The kinetic energy $K$ of the electron is given by:

$$K = \frac{1}{2} mv^2$$

From the centripetal force equation:

$$\frac{mv^2}{r} = \frac{ke^2}{r^2}$$

Solving for $mv^2$ :

$$mv^2 = \frac{ke^2}{r}$$

Therefore, the kinetic energy is:

$$K = \frac{1}{2} \times \frac{ke^2}{r}$$

The potential energy $U$ is:

$$U = - \frac{ke^2}{r}$$

Therefore, the total energy $E$ is:

$$E = K + U = \frac{1}{2} \times \frac{ke^2}{r} - \frac{ke^2}{r} = - \frac{ke^2}{2r}$$

Using the expression for $r$ from the previous question:

$$E = - \frac{ke^2}{2} \times \frac{k e^2 n^2}{4\pi^2 m h^2}$$

Simplifying:

$$E = - \frac{2\pi^2 m h^2 k^2 e^4}{n^2 h^2}$$

Hence, the energy of the electron in the nth orbit is:

$$E_n = - \frac{13.6 \, \text{eV}}{n^2}$$

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4. Question:

What is the energy of an electron in the second orbit of the hydrogen atom?

Solution: From the previous derivation, the energy of an electron in the nth orbit is:

$$E_n = - \frac{13.6 \, \text{eV}}{n^2}$$

For $n = 2$ :

$$E_2 = - \frac{13.6 \, \text{eV}}{2^2} = - \frac{13.6 \, \text{eV}}{4} = - 3.4 \, \text{eV}$$

Hence, the energy of the electron in the second orbit is $-3.4 \, \text{eV}$ .

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5. Question:

Using Bohr’s model, calculate the frequency of the radiation emitted when an electron in the hydrogen atom transitions from the third orbit to the first orbit.

Solution: The energy of the electron in the nth orbit is given by:

$$E_n = - \frac{13.6 \, \text{eV}}{n^2}$$

For the third orbit ($n = 3$ ):

$$E_3 = - \frac{13.6 \, \text{eV}}{3^2} = - 1.51 \, \text{eV}$$

For the first orbit ($n = 1$ ):

$$E_1 = - 13.6 \, \text{eV}$$

The energy difference between the two orbits is:

$$\Delta E = E_1 - E_3 = - 13.6 + 1.51 = - 12.09 \, \text{eV}$$

The frequency of the emitted radiation is related to the energy difference by:

$$E = h f$$

Solving for $f$ :

$$f = \frac{\Delta E}{h}$$

Converting energy to joules:

$$\Delta E = -12.09 \, \text{eV} = -12.09 \times 1.6 \times 10^{-19} \, \text{J} = -1.934 \times 10^{-18} \, \text{J}$$

Now, solving for the frequency:

$$f = \frac{1.934 \times 10^{-18}}{6.626 \times 10^{-34}} = 2.92 \times 10^{15} \, \text{Hz}$$

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6. Question:

If the radius of the hydrogen atom in the first orbit is $5.29 \times 10^{-11} \, \text{m}$ , calculate the radius in the fourth orbit.

Solution: The radius of the nth orbit is given by:

$$r_n = \frac{n^2 h^2}{4\pi^2 m e^2}$$

Since the radius is proportional to $n^2$ , for the fourth orbit:

$$r_4 = r_1 \times \left( \frac{4^2}{1^2} \right) = 5.29 \times 10^{-11} \times 16 = 8.464 \times 10^{-10} \, \text{m}$$

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7. Question:

Explain how the Bohr model of the atom explains the line spectrum of hydrogen.

Solution: The Bohr model explains the line spectrum of hydrogen by suggesting that electrons in a hydrogen atom occupy discrete orbits with quantized energy levels. When an electron jumps from a higher orbit to a lower orbit, it emits a photon with energy equal to the difference between the two orbits. The frequency of the emitted radiation is given by:

$$f = \frac{\Delta E}{h}$$

These discrete transitions result in a series of lines in the spectrum, corresponding to the different energy levels. This explains the discrete nature of the spectral lines observed in hydrogen’s emission spectrum.

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8. Question:

What is the frequency of radiation emitted when an electron in the hydrogen atom transitions from the n = 3 state to the n = 2 state?

Solution: The energy levels are given by:

$$E_n = - \frac{13.6 \, \text{eV}}{n^2}$$

For $n = 3$ :

$$E_3 = - \frac{13.6}{9} = -1.51 \, \text{eV}$$

For $n = 2$ :

$$E_2 = - \frac{13.6}{4} = -3.4 \, \text{eV}$$

The energy difference is:

$$\Delta E = E_2 - E_3 = -3.4 + 1.51 = -1.89 \, \text{eV}$$

Converting to joules:

$$\Delta E = -1.89 \times 1.6 \times 10^{-19} = -3.024 \times 10^{-19} \, \text{J}$$

The frequency of the emitted radiation is:

$$f = \frac{\Delta E}{h} = \frac{3.024 \times 10^{-19}}{6.626 \times 10^{-34}} = 4.57 \times 10^{14} \, \text{Hz}$$

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9. Question:

Explain the concept of spectral series in the hydrogen atom with an example.

Solution: Spectral series in the hydrogen atom are sets of spectral lines that correspond to transitions between different energy levels. The main series in the hydrogen atom are the Lyman, Balmer, Paschen, and Brackett series, named after their discoverers. For example, the Balmer series corresponds to transitions from higher levels ($n = 3, 4, 5, \ldots$ ) to $n = 2$ . These transitions result in visible light emissions. The wavelengths and frequencies of these spectral lines can be calculated using the Rydberg formula.

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10. Question:

What is the significance of the Rydberg constant in atomic physics?

Solution: The Rydberg constant is a fundamental physical constant used to predict the wavelengths of spectral lines in the hydrogen atom. It is part of the Rydberg formula:

$$\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$

where $R_H$ is the Rydberg constant, $n_1$ and $n_2$ are integers representing different energy levels, and $\lambda$ is the wavelength of the emitted or absorbed radiation. The Rydberg constant helps quantify the energy differences between electron orbits in the hydrogen atom.