Question 1

Q1. Choose the correct alternative from the options given below:
(a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no

Accepted Answer

(a) The size of the atom in Thomson’s model is **no different from** the atomic size in Rutherford’s model.

**Explanation:** Both Thomson's and Rutherford's models estimate the atomic size to be approximately 10⁻¹⁰ m (of the order of an angstrom). While their internal structures differ significantly, their overall predicted size for the atom is comparable.

(b) In the ground state of **Thomson’s model** electrons are in stable equilibrium, while in **Rutherford’s model** electrons always experience a net force.

**Explanation:**
*   **Thomson's Model:** Electrons are embedded in a uniform sphere of positive charge. In the ground state, they are at rest or oscillating about equilibrium positions, where the attractive force from the positive sphere balances the repulsive forces from other electrons. Thus, they are in stable equilibrium.
*   **Rutherford's Model:** Electrons revolve around a positively charged nucleus. For an electron to remain in orbit, it must continuously experience a centripetal force (provided by the electrostatic attraction to the nucleus). This centripetal force is a net force, constantly changing the direction of the electron's velocity. Without this net force, the electron would fly off tangentially.

(c) A classical atom based on **Rutherford’s model** is doomed to collapse.

**Explanation:** According to classical electromagnetic theory, an electron revolving in a circular orbit (accelerating) should continuously radiate energy. As it loses energy, its orbit would continuously shrink, and it would eventually spiral into the nucleus, leading to the collapse of the atom. This is a major drawback of Rutherford's model when viewed from classical physics.

(d) The positively charged part of the atom possesses most of the mass in **both the models**.

**Explanation:**
*   **Thomson's Model:** The positive charge is spread throughout the atom, and since electrons have negligible mass, the positive 'pudding' constitutes most of the atom's mass.
*   **Rutherford's Model:** The positive charge is concentrated in a tiny, dense nucleus, which also contains almost all the mass of the atom (protons and neutrons, while electrons have negligible mass). The alpha-particle scattering experiment confirmed this concentration of mass and positive charge.

Question 2

Q2. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Accepted Answer

When a 12.5 eV electron beam bombards gaseous hydrogen at room temperature, the hydrogen atoms are initially in their ground state (n=1). The electrons in the beam transfer energy to the hydrogen atoms, exciting them to higher energy levels. We need to determine the maximum energy level to which the hydrogen atom can be excited and then identify the possible transitions and corresponding spectral series.

Question 3

Q3. What is the minimum energy required to remove an electron from the ground state of a singly ionised helium atom?

Accepted Answer

To remove an electron from the ground state of an atom, we need to provide energy equal to its ionization energy. For a singly ionized helium atom (He⁺), it is a hydrogen-like atom with atomic number Z=2. We can use Bohr's model energy formula for hydrogen-like atoms.

Question 4

Q4. According to the classical electromagnetic theory, calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom.
Assume that the

Accepted Answer

According to classical electromagnetic theory, an electron revolving in a circular orbit is an accelerating charge and should continuously radiate energy. The frequency of the emitted light would be equal to the frequency of revolution of the electron. We will calculate this frequency using classical mechanics and electrostatics.

Question 5

Q5. A hydrogen atom initially in the ground state absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of the photon.

Accepted Answer

When a hydrogen atom absorbs a photon and gets excited from the ground state (n=1) to the n=4 level, the energy of the absorbed photon must be equal to the energy difference between these two levels. We will calculate this energy difference and then use it to find the wavelength and frequency of the photon.

Question 6

Q6. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Accepted Answer

When a 12.5 eV electron beam bombards gaseous hydrogen at room temperature, the hydrogen atoms are initially in the ground state (n=1). The incident electrons can excite the hydrogen atoms to higher energy levels if their energy matches the energy difference between the ground state and an excited state.

We use the energy level formula for hydrogen: E_n = -13.6/n² eV.

1.  **Energy of ground state (n=1):** E₁ = -13.6/1² = -13.6 eV
2.  **Energy of first excited state (n=2):** E₂ = -13.6/2² = -3.4 eV
3.  **Energy of second excited state (n=3):** E₃ = -13.6/3² = -1.51 eV
4.  **Energy of third excited state (n=4):** E₄ = -13.6/4² = -0.85 eV

**Possible excitation energies from ground state (n=1):**
*   To n=2: ΔE₁₂ = E₂ - E₁ = -3.4 eV - (-13.6 eV) = 10.2 eV
*   To n=3: ΔE₁₃ = E₃ - E₁ = -1.51 eV - (-13.6 eV) = 12.09 eV
*   To n=4: ΔE₁₄ = E₄ - E₁ = -0.85 eV - (-13.6 eV) = 12.75 eV

The incident electron beam has an energy of 12.5 eV. This energy is sufficient to excite hydrogen atoms from n=1 to n=2 (10.2 eV) and from n=1 to n=3 (12.09 eV). However, it is not enough to excite them to n=4 (12.75 eV).

Therefore, the hydrogen atoms can be excited to the n=2 and n=3 energy levels.

When these excited atoms de-excite, they emit photons. The possible transitions are:

1.  **From n=3 to n=1:** This transition belongs to the Lyman series.
    Energy: E₃₁ = E₃ - E₁ = -1.51 - (-13.6) = 12.09 eV
    Wavelength: λ = hc/E = (6.626 × 10⁻³⁴ J s × 3 × 10⁸ m/s) / (12.09 eV × 1.602 × 10⁻¹⁹ J/eV) ≈ 102.6 nm

2.  **From n=3 to n=2:** This transition belongs to the Balmer series.
    Energy: E₃₂ = E₃ - E₂ = -1.51 - (-3.4) = 1.89 eV
    Wavelength: λ = hc/E = (6.626 × 10⁻³⁴ J s × 3 × 10⁸ m/s) / (1.89 eV × 1.602 × 10⁻¹⁹ J/eV) ≈ 656.3 nm (H-alpha line)

3.  **From n=2 to n=1:** This transition belongs to the Lyman series.
    Energy: E₂₁ = E₂ - E₁ = -3.4 - (-13.6) = 10.2 eV
    Wavelength: λ = hc/E = (6.626 × 10⁻³⁴ J s × 3 × 10⁸ m/s) / (10.2 eV × 1.602 × 10⁻¹⁹ J/eV) ≈ 121.6 nm

Thus, the emitted wavelengths will correspond to transitions within the **Lyman series** (n=3 to n=1, n=2 to n=1) and the **Balmer series** (n=3 to n=2).

Question 7

Q7. In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴

Accepted Answer

According to Bohr's quantization condition for angular momentum, the angular momentum (L) of an electron in a stable orbit must be an integral multiple of ħ (h/2π).
L = nħ, where n is the principal quantum number and ħ = h/2π.

For a classical object like the Earth revolving around the Sun, its angular momentum is given by L = mvr, where m is the mass, v is the orbital speed, and r is the orbital radius.

Given values:
Mass of Earth (m) = 6.0 × 10²⁴ kg
Orbital radius (r) = 1.5 × 10¹¹ m
Orbital speed (v) = 3 × 10⁴ m/s
Planck's constant (h) = 6.626 × 10⁻³⁴ J s

**Step 1: Calculate the angular momentum of the Earth.**
L = mvr
L = (6.0 × 10²⁴ kg) × (3 × 10⁴ m/s) × (1.5 × 10¹¹ m)
L = 27 × 10²⁴⁺⁴⁺¹¹ kg m²/s
L = 27 × 10³⁹ kg m²/s
L = 2.7 × 10⁴⁰ J s

**Step 2: Apply Bohr's quantization condition.**
L = nħ = n(h/2π)

**Step 3: Solve for the quantum number n.**
n = L / (h/2π)
n = (2πL) / h
n = (2 × 3.14159 × 2.7 × 10⁴⁰ J s) / (6.626 × 10⁻³⁴ J s)
n = (16.9646 × 10⁴⁰) / (6.626 × 10⁻³⁴)
n = 2.560 × 10⁴⁰⁺³⁴
n = 2.560 × 10⁷⁴

Thus, the quantum number that characterises the Earth's revolution around the Sun is approximately **2.56 × 10⁷⁴**.

Question 8

Q8. Choose the correct alternative from the given ones.
(a) Size of the atom in Thomson’s model is …… the atomic size in Rutherford’s model.
(much greater than/no different from/much

Accepted Answer

(a) Size of the atom in Thomson’s model is **no different from** the atomic size in Rutherford’s model.

(b) In the ground state of **Thomson’s model** electrons are in stable equilibrium, while in **Rutherford’s model** electrons always experience a net force.

(c) A classical atom based on Rutherford’s model is **unstable**.

(d) An atom has a **minimum energy** ground state.

(e) The spectral series of hydrogen atom lie in the **ultraviolet, visible, infrared** region of the electromagnetic spectrum (corresponding to Lyman, Balmer, Paschen series).

Question 9

Q9. Does the Bohr’s model of hydrogen atom also apply to two-electron atoms? Explain.

Accepted Answer

No, the Bohr's model of the hydrogen atom does not apply to two-electron atoms (or any multi-electron atoms).

**Explanation:**

1.  **Electron-electron repulsion:** Bohr's model is based on the assumption of a single electron orbiting a stationary nucleus. It does not account for the electrostatic repulsion between multiple electrons in an atom. In a two-electron atom (like Helium), the interaction between the two electrons significantly affects their energy levels and orbital dynamics, which Bohr's model cannot describe.

2.  **Screening effect:** In multi-electron atoms, the inner electrons screen the nuclear charge from the outer electrons. This means the effective nuclear charge experienced by an outer electron is less than the actual nuclear charge. Bohr's model assumes a fixed nuclear charge (Z) and does not incorporate this screening effect.

3.  **Complex interactions:** The presence of multiple electrons leads to much more complex interactions (electron-electron, electron-nucleus, and spin-orbit coupling) that cannot be simplified to a single electron orbiting a nucleus under a simple Coulomb force. Quantum mechanics is required to accurately describe such systems.

Bohr's model is successful only for hydrogen and hydrogen-like ions (single-electron species like He⁺, Li²⁺) where there is only one electron interacting with the nucleus.

Question 10

Q10. A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to elec

Accepted Answer

According to Bohr's angular momentum postulate, the angular momentum (L) of an orbiting body is quantized and given by L = nħ, where n is the principal quantum number and ħ = h/2π.
For a classical object like a satellite in a circular orbit, its angular momentum is given by L = mvr, where m is the mass, v is the orbital speed, and r is the orbital radius.

Given values:
Mass of satellite (m) = 10 kg
Orbital radius (r) = 8000 km = 8000 × 10³ m = 8 × 10⁶ m
Time period (T) = 2 h = 2 × 60 × 60 s = 7200 s
Planck's constant (h) = 6.63 × 10⁻³⁴ J s

**Step 1: Calculate the orbital speed (v) of the satellite.**
The satellite completes one revolution (2πr) in time T.
v = (2πr) / T
v = (2 × 3.14159 × 8 × 10⁶ m) / (7200 s)
v = (50.265 × 10⁶) / 7200 m/s
v ≈ 6981.25 m/s

**Step 2: Calculate the angular momentum (L) of the satellite.**
L = mvr
L = (10 kg) × (6981.25 m/s) × (8 × 10⁶ m)
L = 69812.5 × 8 × 10⁶ kg m²/s
L = 558500 × 10⁶ J s
L = 5.585 × 10¹¹ J s

**Step 3: Apply Bohr's quantization condition and solve for n.**
L = nħ = n(h/2π)
n = L / (h/2π)
n = (2πL) / h
n = (2 × 3.14159 × 5.585 × 10¹¹ J s) / (6.63 × 10⁻³⁴ J s)
n = (35.107 × 10¹¹) / (6.63 × 10⁻³⁴)
n = 5.295 × 10¹¹⁺³⁴
n = 5.295 × 10⁴⁵

Thus, the quantum number of the orbit of the satellite is approximately **5.30 × 10⁴⁵**.

Question 11

Q11. Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.

Accepted Answer

The wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum are:
1.  For n₂ = 2 (first line): λ₁ = 121.5 nm
2.  For n₂ = 3 (second line): λ₂ = 102.5 nm
3.  For n₂ = 4 (third line): λ₃ = 97.2 nm
4.  For n₂ = 5 (fourth line): λ₄ = 94.9 nm

Question 12

Q12. What is the shortest wavelength present in the Paschen series of the hydrogen spectrum?

Accepted Answer

The shortest wavelength present in the Paschen series of the hydrogen spectrum is 820.4 nm.

Question 13

Q13. A hydrogen atom initially in its ground state absorbs a photon which excites it to the n = 4 level. Determine the wavelength and frequency of the photon.

Accepted Answer

The wavelength of the absorbed photon is 97.2 nm and its frequency is 3.08 × 10¹⁵ Hz.

Question 14

Q14. (a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.

Accepted Answer

(a) The speeds of the electron in a hydrogen atom are:
   - For n = 1: v₁ = 2.19 × 10⁶ m/s
   - For n = 2: v₂ = 1.09 × 10⁶ m/s
   - For n = 3: v₃ = 7.29 × 10⁵ m/s
(b) The orbital periods in these levels are:
   - For n = 1: T₁ = 1.52 × 10⁻¹⁶ s
   - For n = 2: T₂ = 1.22 × 10⁻¹⁵ s
   - For n = 3: T₃ = 4.10 × 10⁻¹⁵ s

Question 15

Q15. The spectral line of hydrogen atom in the Balmer series has wavelength 656.3 nm. What is the transition responsible for this radiation? (R = 1.097 × 10⁷ m⁻¹)

Accepted Answer

The transition responsible for this radiation is from n₂ = 3 to n₁ = 2.

Question 16

Q16. A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of the photon.

Accepted Answer

The hydrogen atom is initially in the ground state (n=1) and is excited to the n=4 level. The energy absorbed by the photon is equal to the difference in energy between the n=4 and n=1 levels. Using this energy, the wavelength and frequency of the photon can be determined.

**Calculations:**
1.  **Energy of the ground state (n=1):**
    E₁ = -13.6 eV
2.  **Energy of the excited state (n=4):**
    E₄ = -13.6 / n² = -13.6 / 4² = -13.6 / 16 = -0.85 eV
3.  **Energy absorbed by the photon (ΔE):**
    ΔE = E₄ - E₁ = -0.85 eV - (-13.6 eV) = 12.75 eV
4.  **Convert energy to Joules:**
    ΔE = 12.75 eV × (1.602 × 10⁻¹⁹ J/eV) = 2.04255 × 10⁻¹⁸ J
5.  **Calculate the wavelength (λ) of the photon:**
    Using the relation ΔE = hc/λ, where h = 6.626 × 10⁻³⁴ J s (Planck's constant) and c = 3 × 10⁸ m/s (speed of light).
    λ = hc / ΔE
    λ = (6.626 × 10⁻³⁴ J s × 3 × 10⁸ m/s) / (2.04255 × 10⁻¹⁸ J)
    λ = (1.9878 × 10⁻²⁵ J m) / (2.04255 × 10⁻¹⁸ J)
    λ ≈ 9.732 × 10⁻⁸ m = 97.32 nm
6.  **Calculate the frequency (ν) of the photon:**
    Using the relation ΔE = hν, or ν = c/λ.
    ν = ΔE / h
    ν = (2.04255 × 10⁻¹⁸ J) / (6.626 × 10⁻³⁴ J s)
    ν ≈ 3.083 × 10¹⁵ Hz

**Final Answer:**
The wavelength of the photon is approximately **97.32 nm**.
The frequency of the photon is approximately **3.083 × 10¹⁵ Hz**.

Question 17

Q17. Using the Bohr model, calculate the speed and period of revolution of an electron in the n = 2 orbit of a hydrogen atom.

Accepted Answer

According to Bohr's model, the speed of an electron in the nth orbit of a hydrogen atom is given by v_n = (cα)/n, where c is the speed of light and α is the fine-structure constant. The radius of the nth orbit is given by r_n = n²a₀, where a₀ is the Bohr radius. The period of revolution can then be calculated using T_n = 2πr_n / v_n.

**Given values:**
*   n = 2 (for the second orbit)
*   Speed of light, c = 3 × 10⁸ m/s
*   Fine-structure constant, α = e² / (2ε₀hc) ≈ 1/137
*   Bohr radius, a₀ = 0.529 × 10⁻¹⁰ m

**Calculations:**
1.  **Calculate the speed (v₂) of the electron in the n=2 orbit:**
    v_n = (cα) / n
    v₂ = (3 × 10⁸ m/s × (1/137)) / 2
    v₂ = (3 × 10⁸ m/s) / 274
    v₂ ≈ 1.095 × 10⁶ m/s

Alternatively, using the formula v_n = (e² / (2ε₀h)) * (1/n) = (2.18 × 10⁶ m/s) / n
    v₂ = (2.18 × 10⁶ m/s) / 2
    v₂ = 1.09 × 10⁶ m/s

2.  **Calculate the radius (r₂) of the n=2 orbit:**
    r_n = n²a₀
    r₂ = 2² × (0.529 × 10⁻¹⁰ m)
    r₂ = 4 × 0.529 × 10⁻¹⁰ m
    r₂ = 2.116 × 10⁻¹⁰ m

3.  **Calculate the period of revolution (T₂) of the electron:**
    T_n = 2πr_n / v_n
    T₂ = (2π × 2.116 × 10⁻¹⁰ m) / (1.095 × 10⁶ m/s)
    T₂ = (13.294 × 10⁻¹⁰ m) / (1.095 × 10⁶ m/s)
    T₂ ≈ 1.214 × 10⁻¹⁵ s

**Final Answer:**
The speed of the electron in the n=2 orbit is approximately **1.095 × 10⁶ m/s**.
The period of revolution of the electron in the n=2 orbit is approximately **1.214 × 10⁻¹⁵ s**.

Question 18

Q18. The spectral series of the hydrogen atom are given by the formula: 1/λ = R (1/n_f^2 - 1/n_i^2) where R is the Rydberg constant. Calculate the shortest wavelength of the Balmer seri

Accepted Answer

The Rydberg formula describes the wavelengths of spectral lines for hydrogen. For the Balmer series, the final energy level (n_f) is 2. The shortest wavelength corresponds to the highest energy transition, which occurs when the electron falls from an infinitely high energy level (n_i = ∞) to the n_f = 2 level.

**Given values:**
*   Rydberg constant, R = 1.097 × 10⁷ m⁻¹
*   For Balmer series, n_f = 2
*   For shortest wavelength, n_i = ∞

**Calculations:**
1.  **Apply the Rydberg formula:**
    1/λ = R (1/n_f² - 1/n_i²)
2.  **Substitute values for Balmer series and shortest wavelength:**
    1/λ = R (1/2² - 1/∞²)
    1/λ = R (1/4 - 0)
    1/λ = R/4
3.  **Calculate the wavelength (λ):**
    λ = 4/R
    λ = 4 / (1.097 × 10⁷ m⁻¹)
    λ ≈ 3.646 × 10⁻⁷ m
    λ = 364.6 nm

**Final Answer:**
The shortest wavelength of the Balmer series is approximately **364.6 nm**.

Question 19

Q19. A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of the photon.

Accepted Answer

The hydrogen atom transitions from the ground state (n=1) to the n=4 level. The energy difference between these levels corresponds to the energy of the absorbed photon. Using this energy, we can calculate the frequency and wavelength of the photon.

**Calculations:**
1.  **Energy of the photon (ΔE):**
    The energy of an electron in the nth orbit of a hydrogen atom is given by E_n = -13.6/n^2 eV.
    Initial energy level, n_i = 1 (ground state).
    Final energy level, n_f = 4.
    E_1 = -13.6/1^2 = -13.6 eV
    E_4 = -13.6/4^2 = -13.6/16 = -0.85 eV
    Energy absorbed by the photon, ΔE = E_f - E_i = E_4 - E_1 = -0.85 eV - (-13.6 eV) = 12.75 eV.
    Convert energy to Joules: ΔE = 12.75 eV * (1.602 x 10^-19 J/eV) = 2.04255 x 10^-18 J.

2.  **Frequency of the photon (ν):**
    The energy of a photon is given by ΔE = hν, where h is Planck's constant (6.626 x 10^-34 J s).
    ν = ΔE / h = (2.04255 x 10^-18 J) / (6.626 x 10^-34 J s) = 3.0826 x 10^15 Hz.

3.  **Wavelength of the photon (λ):**
    The relationship between frequency, wavelength, and the speed of light (c = 3 x 10^8 m/s) is c = λν.
    λ = c / ν = (3 x 10^8 m/s) / (3.0826 x 10^15 Hz) = 9.732 x 10^-8 m.
    λ = 97.32 nm.

Therefore, the wavelength of the photon is approximately **97.32 nm** and its frequency is approximately **3.08 x 10^15 Hz**.

Question 20

Q20. Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. Also calculate the orbital period in each of these levels.

Accepted Answer

According to the Bohr model, the speed of an electron in the nth orbit of a hydrogen-like atom (with atomic number Z) is given by:
v_n = (Z e^2) / (2 ε_0 n h)
For a hydrogen atom, Z = 1. So, v_n = e^2 / (2 ε_0 n h)

And the orbital radius is given by:
r_n = (n^2 h^2 ε_0) / (π m_e e^2 Z)
For hydrogen, Z = 1. So, r_n = (n^2 h^2 ε_0) / (π m_e e^2) = n^2 a_0, where a_0 is the Bohr radius (a_0 ≈ 0.529 x 10^-10 m).

The orbital period (T_n) is the time taken for one complete revolution, which is given by:
T_n = 2πr_n / v_n

**Constants:**
e = 1.602 x 10^-19 C
ε_0 = 8.854 x 10^-12 C^2 N^-1 m^-2
h = 6.626 x 10^-34 J s
m_e = 9.109 x 10^-31 kg
π = 3.14159

**Calculations for Speed (v_n):**
Let's first calculate the constant term: e^2 / (2 ε_0 h)
= (1.602 x 10^-19 C)^2 / (2 * 8.854 x 10^-12 C^2 N^-1 m^-2 * 6.626 x 10^-34 J s)
= (2.5664 x 10^-38) / (1.173 x 10^-44)
≈ 2.188 x 10^6 m/s
This is approximately c/137, where c is the speed of light (fine structure constant α = v_1/c ≈ 1/137).
So, v_n = (2.188 x 10^6 m/s) / n

*   **For n = 1:**
    v_1 = (2.188 x 10^6 m/s) / 1 = **2.188 x 10^6 m/s**
*   **For n = 2:**
    v_2 = (2.188 x 10^6 m/s) / 2 = **1.094 x 10^6 m/s**
*   **For n = 3:**
    v_3 = (2.188 x 10^6 m/s) / 3 = **7.293 x 10^5 m/s**

**Calculations for Orbital Period (T_n):**
First, calculate the orbital radius r_n = n^2 a_0, where a_0 = 0.529 x 10^-10 m.

*   **For n = 1:**
    r_1 = 1^2 * (0.529 x 10^-10 m) = 0.529 x 10^-10 m
    T_1 = 2πr_1 / v_1 = (2 * 3.14159 * 0.529 x 10^-10 m) / (2.188 x 10^6 m/s)
    T_1 = (3.323 x 10^-10) / (2.188 x 10^6) = **1.519 x 10^-16 s**

*   **For n = 2:**
    r_2 = 2^2 * (0.529 x 10^-10 m) = 4 * 0.529 x 10^-10 m = 2.116 x 10^-10 m
    T_2 = 2πr_2 / v_2 = (2 * 3.14159 * 2.116 x 10^-10 m) / (1.094 x 10^6 m/s)
    T_2 = (1.329 x 10^-9) / (1.094 x 10^6) = **1.215 x 10^-15 s**

*   **For n = 3:**
    r_3 = 3^2 * (0.529 x 10^-10 m) = 9 * 0.529 x 10^-10 m = 4.761 x 10^-10 m
    T_3 = 2πr_3 / v_3 = (2 * 3.14159 * 4.761 x 10^-10 m) / (7.293 x 10^5 m/s)
    T_3 = (2.991 x 10^-9) / (7.293 x 10^5) = **4.101 x 10^-15 s**

**Summary:**
| Level (n) | Speed (v_n) (m/s) | Orbital Period (T_n) (s) |
| :-------- | :------------------ | :----------------------- |
| 1         | 2.188 x 10^6        | 1.519 x 10^-16           |
| 2         | 1.094 x 10^6        | 1.215 x 10^-15           |
| 3         | 7.293 x 10^5        | 4.101 x 10^-15           |

Question 21

Q21. The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?

Accepted Answer

For a hydrogen atom in its ground state, the total energy (E) is -13.6 eV. According to Bohr's model, there is a specific relationship between the total energy, kinetic energy (K), and potential energy (U) of the electron.

1.  **Kinetic Energy (K):**
    The total energy of an electron in a stationary orbit is the sum of its kinetic and potential energies. For a hydrogen-like atom, the total energy (E) is related to the kinetic energy (K) by the formula: E = -K.
    Given E = -13.6 eV.
    Therefore, K = -E = -(-13.6 eV) = +13.6 eV.

2.  **Potential Energy (U):**
    The total energy (E) is also related to the potential energy (U) by the formula: E = U/2.
    Therefore, U = 2E = 2 × (-13.6 eV) = -27.2 eV.

Alternatively, since E = K + U, we can find U as:
U = E - K = -13.6 eV - 13.6 eV = -27.2 eV.

Thus, in the ground state of the hydrogen atom, the kinetic energy of the electron is +13.6 eV, and its potential energy is -27.2 eV.

Question 22

Q22. A hydrogen atom initially in its ground state absorbs a photon which excites it to the n = 4 level. Determine the wavelength and frequency of the photon.

Accepted Answer

To determine the wavelength and frequency of the absorbed photon, we first need to calculate the energy difference between the initial (ground) state and the final (n=4) excited state.

**1. Energy of the electron in different states:**
The energy of an electron in the n-th orbit of a hydrogen atom is given by the formula:
E_n = -13.6 / n^2 eV

*   **Ground state (n=1):**
    E_1 = -13.6 / 1^2 eV = -13.6 eV

*   **Excited state (n=4):**
    E_4 = -13.6 / 4^2 eV = -13.6 / 16 eV = -0.85 eV

**2. Energy of the absorbed photon (ΔE):**
The energy of the absorbed photon is equal to the energy difference between the final and initial states:
ΔE = E_4 - E_1
ΔE = (-0.85 eV) - (-13.6 eV)
ΔE = 12.75 eV

**3. Convert energy to Joules:**
Since 1 eV = 1.602 × 10^-19 J,
ΔE = 12.75 eV × (1.602 × 10^-19 J/eV)
ΔE = 2.04255 × 10^-18 J

**4. Calculate the frequency (f) of the photon:**
The energy of a photon is given by Planck's equation: ΔE = hf
Where h is Planck's constant (6.626 × 10^-34 J s).

f = ΔE / h
f = (2.04255 × 10^-18 J) / (6.626 × 10^-34 J s)
f ≈ 3.082 × 10^15 Hz

**5. Calculate the wavelength (λ) of the photon:**
The relationship between frequency, wavelength, and the speed of light (c) is: c = fλ
Where c is the speed of light (3 × 10^8 m/s).

λ = c / f
λ = (3 × 10^8 m/s) / (3.082 × 10^15 Hz)
λ ≈ 9.734 × 10^-8 m

Converting to nanometers (1 nm = 10^-9 m):
λ ≈ 97.34 nm

**Summary:**
The frequency of the photon is approximately 3.082 × 10^15 Hz.
The wavelength of the photon is approximately 9.734 × 10^-8 m (or 97.34 nm).

Question 23

Q23. Using the Bohr model, calculate the speed and period of revolution of the electron in the n = 1 orbit of a hydrogen atom. (Take k = 9 × 10^9 Nm^2 C^–2, e = 1.6 × 10^–19 C, m_e = 9.

Accepted Answer

To calculate the speed and period of revolution of the electron in the n=1 orbit of a hydrogen atom, we will use the fundamental principles of Bohr's model.

**Given constants:**
*   Coulomb's constant, k = 9 × 10^9 Nm^2 C^-2
*   Charge of electron, e = 1.6 × 10^-19 C
*   Mass of electron, m_e = 9.1 × 10^-31 kg
*   Planck's constant, h = 6.6 × 10^-34 J s
*   For hydrogen atom, atomic number Z = 1
*   For ground state, principal quantum number n = 1

**1. Calculate the speed (v_n) of the electron:**
According to Bohr's model, the speed of an electron in the n-th orbit of a hydrogen-like atom is given by:
v_n = (Z e^2) / (2 n h ε_0)
We know that k = 1 / (4π ε_0), so ε_0 = 1 / (4πk).
Substituting ε_0 into the speed formula:
v_n = (Z e^2) / (2 n h * (1 / (4πk)))
v_n = (2πk Z e^2) / (n h)

For n=1 and Z=1 (hydrogen atom ground state):
v_1 = (2π × (9 × 10^9 Nm^2 C^-2) × (1) × (1.6 × 10^-19 C)^2) / (1 × 6.6 × 10^-34 J s)
v_1 = (2 × 3.14159 × 9 × 10^9 × 2.56 × 10^-38) / (6.6 × 10^-34)
v_1 = (144.764 × 10^-29) / (6.6 × 10^-34)
v_1 ≈ 21.934 × 10^4 m/s
v_1 ≈ 2.19 × 10^6 m/s

**2. Calculate the radius (r_n) of the electron's orbit:**
The radius of the n-th orbit of a hydrogen-like atom is given by:
r_n = (n^2 h^2 ε_0) / (π m_e Z e^2)
Substituting ε_0 = 1 / (4πk):
r_n = (n^2 h^2) / (π m_e Z e^2 * 4πk)
r_n = (n^2 h^2) / (4π^2 m_e Z e^2 k)

For n=1 and Z=1 (hydrogen atom ground state):
r_1 = (1^2 × (6.6 × 10^-34 J s)^2) / (4π^2 × (9.1 × 10^-31 kg) × (1) × (1.6 × 10^-19 C)^2 × (9 × 10^9 Nm^2 C^-2))
r_1 = (43.56 × 10^-68) / (4 × (3.14159)^2 × 9.1 × 10^-31 × 2.56 × 10^-38 × 9 × 10^9)
r_1 = (43.56 × 10^-68) / (4 × 9.8696 × 9.1 × 2.56 × 9 × 10^-31-38+9)
r_1 = (43.56 × 10^-68) / (8304.7 × 10^-60)
r_1 ≈ 0.005245 × 10^-8 m
r_1 ≈ 5.29 × 10^-11 m

This is the Bohr radius (a_0), which is approximately 0.529 Å or 5.29 × 10^-11 m.

**3. Calculate the period of revolution (T_n):**
The period of revolution is the time taken for one complete orbit, which is the circumference divided by the speed:
T_n = (2πr_n) / v_n

For n=1:
T_1 = (2π × 5.29 × 10^-11 m) / (2.19 × 10^6 m/s)
T_1 = (2 × 3.14159 × 5.29 × 10^-11) / (2.19 × 10^6)
T_1 = (33.23 × 10^-11) / (2.19 × 10^6)
T_1 ≈ 15.17 × 10^-17 s
T_1 ≈ 1.52 × 10^-16 s

**Summary:**
*   Speed of the electron in the n=1 orbit: **2.19 × 10^6 m/s**
*   Period of revolution of the electron in the n=1 orbit: **1.52 × 10^-16 s**

Chapter 12: Atoms