AtomsClass 12 Physics NCERT Solutions
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Solution 1 of 9
Q1EXERCISES
Choose the correct alternative from the clues given at the end of the each statement:
(a)
The size of the atom in Thomson's model is ......... the atomic size in Rutherford's model. (much greater than/no different from/much less than.)
(b)
In the ground state of ......... electrons are in stable equilibrium, while in ......... electrons always experience a net force. (Thomson's model/ Rutherford's model.)
(c)
A classical atom based on ......... is doomed to collapse. (Thomson's model/ Rutherford's model.)
(d)
An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson's model/ Rutherford's model.)
(e) The positively charged part of the atom possesses most of the mass in ......... (Rutherford's model/both the models.)
Solution
(a) The size of the atom in Thomson's model is no different from the atomic size in Rutherford's model. Both models estimated the atomic radius to be about 10⁻¹⁰ m.
(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. In Thomson's model, the electrons are embedded in the positive charge and are in equilibrium. In Rutherford's model, the electrons revolve around the nucleus and are constantly under a centripetal force.
(c) A classical atom based on Rutherford's model is doomed to collapse. According to classical electromagnetic theory, an accelerating electron (in a circular orbit) must radiate energy and spiral into the nucleus.
(d) An atom has a nearly continuous mass distribution in a Thomson's model but has a highly non-uniform mass distribution in Rutherford's model. In Thomson's model, the positive charge and mass are distributed uniformly, whereas in Rutherford's model, they are concentrated in a tiny nucleus.
(e) The positively charged part of the atom possesses most of the mass in both the models. In Thomson's model, the mass of the embedded electrons is negligible compared to the positive sphere. In Rutherford's model, the nucleus contains the positive charge and nearly all the mass of the atom.
Q2EXERCISES
Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K .) What results do you expect?
Solution
If the alpha-particle scattering experiment is repeated with a thin sheet of solid hydrogen, the results would be significantly different from the experiment with the gold foil.
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Mass Comparison: A hydrogen nucleus is a single proton, which has a mass of approximately 1.67 × 10⁻²⁷ kg. An alpha particle (a helium nucleus) has a mass of approximately 6.64 × 10⁻²⁷ kg, which is about four times the mass of a proton. In the original experiment, the gold nucleus was about 50 times heavier than the alpha particle, so it could be considered stationary.
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Collision Dynamics: Since the target (proton) is much lighter than the projectile (alpha particle), the proton cannot be assumed to remain stationary during the collision. In a head-on collision, the alpha particle would not rebound back (scatter at ~180°). Instead, it would drive the proton forward, similar to a heavy ball hitting a lighter one. The law of conservation of momentum dictates that the massive alpha particle cannot have its direction of motion reversed by the much lighter proton.
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Expected Observations:
- Most alpha particles would still pass through undeviated, as the hydrogen atom is also mostly empty space.
- The number of particles scattered at large angles would be extremely small, far fewer than in the gold foil experiment. There would be no backscattering of alpha particles.
- The protons in the solid hydrogen would be knocked out with high velocities after collisions.
Q3EXERCISES
A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?
Solution
According to Bohr's third postulate, when an atom makes a transition from a higher energy level to a lower energy level, it emits a photon with energy equal to the difference between the two levels.
Given:
- Energy difference, ΔE = 2.3 eV
First, we convert the energy from electron volts (eV) to joules (J):
- ΔE = 2.3 × (1.6 × 10⁻¹⁹ J) = 3.68 × 10⁻¹⁹ J
The energy of the emitted photon is related to its frequency (ν) by the Planck-Einstein relation:
- ΔE = hν where h is Planck's constant (h ≈ 6.63 × 10⁻³⁴ J s).
Now, we can calculate the frequency of the emitted radiation:
- ν = ΔE / h
- ν = (3.68 × 10⁻¹⁹ J) / (6.63 × 10⁻³⁴ J s)
- ν ≈ 5.55 × 10¹⁴ Hz
Thus, the frequency of the radiation emitted is approximately 5.55 × 10¹⁴ Hz.
Q4EXERCISES
The ground state energy of hydrogen atom is -13.6 eV . What are the kinetic and potential energies of the electron in this state?
Solution
For an electron in a hydrogen atom, the total energy (E) is the sum of its kinetic energy (K) and potential energy (U).
Given:
- Total energy in the ground state, E = -13.6 eV
From the analysis of electron orbits in the Rutherford model (which is also used in the Bohr model), we have the following relationships:
-
Kinetic Energy (K):
K = (1/2)mv² = e² / (8πε₀r)
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Potential Energy (U):
U = -e² / (4πε₀r)
-
Total Energy (E):
E = K + U = (e² / (8πε₀r)) - (e² / (4πε₀r)) = -e² / (8πε₀r)
By comparing the expressions for K and E, we can see that:
K = -E
By comparing the expressions for U and E, we can see that:
U = 2E
Using these relationships for the ground state:
-
Kinetic Energy (K):
- K = -E = -(-13.6 eV) = 13.6 eV
-
Potential Energy (U):
- U = 2E = 2 × (-13.6 eV) = -27.2 eV
So, in the ground state, the kinetic energy of the electron is 13.6 eV and its potential energy is -27.2 eV.
Q5EXERCISES
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 photon.
Solution
The energy of the absorbed photon must be equal to the energy difference between the final state (n = 4) and the initial state (n = 1).
The energy of an electron in the n-th level of a hydrogen atom is given by:
Eₙ = -13.6 / n² eV
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Calculate the energy of the initial and final states:
- Initial state (ground level):
nᵢ = 1E₁ = -13.6 / 1² = -13.6 eV
- Final state:
n = 4E₄ = -13.6 / 4² = -13.6 / 16 = -0.85 eV
- Initial state (ground level):
-
Calculate the energy of the absorbed photon (ΔE):
ΔE = E₄ - E₁ = (-0.85 eV) - (-13.6 eV) = 12.75 eV
-
Determine the frequency (ν) of the photon:
- First, convert the energy to Joules:
ΔE = 12.75 eV × (1.6 × 10⁻¹⁹ J/eV) = 2.04 × 10⁻¹⁸ J
- Using the relation
ΔE = hν(where h = 6.63 × 10⁻³⁴ J s):ν = ΔE / h = (2.04 × 10⁻¹⁸ J) / (6.63 × 10⁻³⁴ J s)ν ≈ 3.08 × 10¹⁵ Hz
- First, convert the energy to Joules:
-
Determine the wavelength (λ) of the photon:
- Using the relation
c = νλ(where c = 3 × 10⁸ m/s):λ = c / ν = (3 × 10⁸ m/s) / (3.08 × 10¹⁵ Hz)λ ≈ 9.74 × 10⁻⁸ m
- This can also be expressed in nanometers:
λ = 97.4 nm
- Using the relation
Therefore, the frequency of the photon is approximately 3.08 × 10¹⁵ Hz and its wavelength is 97.4 nm.
Q6EXERCISES
(a) Using the Bohr's 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.
Solution
(a) Speed of the electron
From Bohr's model, the speed of an electron in the n-th orbit of a hydrogen atom is given by:
vₙ = e² / (2nε₀h)
Substituting the values of the constants:
e = 1.6 × 10⁻¹⁹ C, ε₀ = 8.85 × 10⁻¹² C²/Nm², h = 6.63 × 10⁻³⁴ J s
vₙ = (1.6 × 10⁻¹⁹)² / (2n × 8.85 × 10⁻¹² × 6.63 × 10⁻³⁴)
vₙ ≈ (2.18 × 10⁶) / n m/s- For n = 1:
v₁ = (2.18 × 10⁶) / 1 = 2.18 × 10⁶ m/s - For n = 2:
v₂ = (2.18 × 10⁶) / 2 = 1.09 × 10⁶ m/s - For n = 3:
v₃ = (2.18 × 10⁶) / 3 ≈ 7.27 × 10⁵ m/s
(b) Orbital period of the electron
The orbital period (T) is the time taken to complete one revolution, given by
T = 2πr / v.
The radius of the n-th orbit is given by rₙ = n²r₁, where r₁ (Bohr radius) is 5.3 × 10⁻¹¹ m.
So, Tₙ = (2π(n²r₁)) / vₙ.
Since vₙ = v₁/n, we can write Tₙ = (2πn²r₁) / (v₁/n) = (2πr₁/v₁) × n³.First, let's calculate the period for n=1:
T₁ = (2πr₁) / v₁ = (2 × 3.14 × 5.3 × 10⁻¹¹) / (2.18 × 10⁶) ≈ 1.53 × 10⁻¹⁶ s
Now we can find the periods for n=2 and n=3:
- For n = 1:
T₁ = 1.53 × 10⁻¹⁶ s - For n = 2:
T₂ = T₁ × 2³ = (1.53 × 10⁻¹⁶) × 8 = 1.224 × 10⁻¹⁵ s - For n = 3:
T₃ = T₁ × 3³ = (1.53 × 10⁻¹⁶) × 27 ≈ 4.13 × 10⁻¹⁵ s
Q7EXERCISES
The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10⁻¹¹ m. What are the radii of the n=2 and n=3 orbits?
Solution
According to Bohr's model for the hydrogen atom, the radius of the n-th stationary orbit is directly proportional to the square of the principal quantum number (n).
The formula for the radius is:
rₙ = n² * r₁
Where:
rₙis the radius of the n-th orbit.nis the principal quantum number.r₁is the radius of the innermost orbit (n=1), also known as the Bohr radius.
Given:
- Radius of the innermost orbit,
r₁ = 5.3 × 10⁻¹¹ m
Radius of the n=2 orbit:
- For n = 2,
r₂ = 2² * r₁ r₂ = 4 × (5.3 × 10⁻¹¹ m)r₂ = 21.2 × 10⁻¹¹ mor2.12 × 10⁻¹⁰ m
Radius of the n=3 orbit:
- For n = 3,
r₃ = 3² * r₁ r₃ = 9 × (5.3 × 10⁻¹¹ m)r₃ = 47.7 × 10⁻¹¹ mor4.77 × 10⁻¹⁰ m
Thus, the radius of the n=2 orbit is 2.12 × 10⁻¹⁰ m, and the radius of the n=3 orbit is 4.77 × 10⁻¹⁰ m.
Q8EXERCISES
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Solution
At room temperature, hydrogen atoms are in their ground state (n=1), which has an energy of E₁ = -13.6 eV.
The bombarding electrons have a kinetic energy of 12.5 eV. A hydrogen atom can be excited to a higher energy level if it absorbs an amount of energy from a colliding electron that is exactly equal to the difference in energy levels.
Let's calculate the energy required to excite the hydrogen atom from the ground state (n=1) to higher states (n=2, 3, 4...):
- To n=2:
ΔE = E₂ - E₁ = (-13.6/2²) - (-13.6) = -3.4 + 13.6 = 10.2 eV - To n=3:
ΔE = E₃ - E₁ = (-13.6/3²) - (-13.6) = -1.51 + 13.6 = 12.09 eV - To n=4:
ΔE = E₄ - E₁ = (-13.6/4²) - (-13.6) = -0.85 + 13.6 = 12.75 eV
Since the incident electrons have 12.5 eV of energy, they can provide enough energy to excite the hydrogen atoms to the n=2 level (requires 10.2 eV) and the n=3 level (requires 12.09 eV). They do not have enough energy to excite the atoms to the n=4 level (requires 12.75 eV).
Therefore, the hydrogen atoms can be excited to the n=2 and n=3 states.
When these excited atoms de-excite, they emit photons. The possible transitions are:
- From n=3 to n=2
- From n=3 to n=1
- From n=2 to n=1
The spectral series are defined by the final state (n) of the transition:
- Lyman Series: Transitions ending at n = 1. (Transitions 2 and 3 above)
- Balmer Series: Transitions ending at n = 2. (Transition 1 above)
Therefore, the emitted radiation will contain wavelengths corresponding to both the Lyman series and the Balmer series.
Q9EXERCISES
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 m/s. (Mass of earth = 6.0 × 10²⁴ kg.)
Solution
According to Bohr's second postulate (the quantization condition), the angular momentum (L) of an orbiting body is an integral multiple of h/2π.
L = mvr = n * (h / 2π)
Where:
mis the mass of the orbiting body (Earth).vis its orbital speed.ris the radius of the orbit.his Planck's constant.nis the principal quantum number.
We need to find the value of
n.Given data:
- Mass of Earth,
m = 6.0 × 10²⁴ kg - Orbital speed,
v = 3 × 10⁴ m/s - Orbital radius,
r = 1.5 × 10¹¹ m - Planck's constant,
h = 6.63 × 10⁻³⁴ J s
Rearranging the formula to solve for
n:n = (2πmvr) / h
Now, substitute the given values into the equation:
n = (2 × 3.14159 × (6.0 × 10²⁴ kg) × (3 × 10⁴ m/s) × (1.5 × 10¹¹ m)) / (6.63 × 10⁻³⁴ J s)n = (1.696 × 10⁴¹) / (6.63 × 10⁻³⁴)n ≈ 2.56 × 10⁷⁴
The quantum number that characterizes the Earth's revolution is approximately 2.56 × 10⁷⁴. This is an extremely large number, which illustrates why quantum effects are not observable for macroscopic objects. The energy levels are so close together that they form a continuum from a classical perspective.