NEET Physics: Atoms questions with solutions

Q1. The ground state energy of H-atom 13.6 eV. The energy needed to ionize H-atom from its second excited state.

1. 1.51 eV
2. 3.4 eV
3. 13.6 eV
4. 12.1 eV

Answer: 3.4 eV

The second excited state corresponds to n=3. The energy of the nth state is given by E_n = -13.6/n² eV. For n=3, E_3 = -13.6/9 = -1.51 eV. To ionize, the energy required is the difference between 0 eV (ionized state) and -1.51 eV, which is 1.51 eV.

Q2. To explain his theory, Bohr used

1. conservation of linear momentum
2. conservation of angular momentum
3. conservation of quantum frequency
4. conservation of energy

Answer: conservation of angular momentum

Bohr used the concept of conservation of angular momentum to explain the quantization of electron orbits in his atomic model.

Q3. The ionisation energy of hydrogen atom is 13.6 eV, the ionisation energy of helium atom would be

1. 13.6 eV
2. 27.2 eV
3. 6.8 eV
4. 54.4 eV

Answer: 54.4 eV

The ionisation energy of a hydrogen-like atom is proportional to the square of its atomic number (Z). For helium (Z=2), the ionisation energy is 13.6 × Z² = 13.6 × 4 = 54.4 eV.

Q4. For the wavelength λ = 975 Å, calculate the number of spectral lines emitted when an electron transitions from n = 4 to n = 1.

1. n(n − 1)/2
2. n(n + 1)/2
3. n(n − 1)
4. n(n + 1)

Answer: n(n − 1)/2

The number of spectral lines emitted during a transition from a higher energy level (n = 4) to a lower energy level (n = 1) is given by the formula n(n − 1)/2, where n is the initial energy level. Substituting n = 4, the number of lines is 4(4 − 1)/2 = 6.

Q5. The ratio of temperatures T₁/T₂ is proportional to the cube of the ratio of principal quantum numbers n₁/n₂. If n₁ = 2n₂, find the ratio T₁/T₂.

1. T₁/T₂ ∝ n₁²/n₂²
2. T₁/T₂ ∝ n₁³/n₂³
3. T₁/T₂ ∝ n₁/n₂
4. T₁/T₂ ∝ n₁⁴/n₂⁴

Answer: T₁/T₂ ∝ n₁³/n₂³

The problem states that T₁/T₂ is proportional to the cube of the ratio of n₁/n₂. Substituting n₁ = 2n₂, we get (n₁/n₂)³ = (2n₂/n₂)³ = 2³ = 8. Thus, T₁/T₂ ∝ n₁³/n₂³.

Q6. According to the Rydberg formula, calculate the wavelength λ₁ for the transition from n = 4 to n = 3.

1. λ₁ = R[1/n₁² − 1/n₂²]
2. λ₁ = R[1/n₂² − 1/n₁²]
3. λ₁ = R[1/n₁ − 1/n₂]
4. λ₁ = R[1/n₂ − 1/n₁]

Answer: λ₁ = R[1/n₂² − 1/n₁²]

The Rydberg formula for the wavelength of light emitted during an electron transition is given by λ₁ = 1/R[1/n₁² − 1/n₂²], where n₁ < n₂. Option B correctly represents this formula.

Q7. For emission, the wave number of the radiation is given as 1/λ = RZ²[1/n₁² − 1/n₂²]. Calculate the wave number for Z = 2, n₁ = 5, and n₂ = 2.

1. 1/λ = RZ²[1/n₁² − 1/n₂²]
2. 1/λ = RZ[1/n₁² − 1/n₂²]
3. 1/λ = R[1/n₁² − 1/n₂²]
4. 1/λ = ZR[1/n₁² − 1/n₂²]

Answer: 1/λ = RZ²[1/n₁² − 1/n₂²]

The formula for the wave number of emitted radiation is correctly given as 1/λ = RZ²[1/n₁² − 1/n₂²]. This is derived from the Rydberg formula for hydrogen-like atoms, where Z is the atomic number, and n₁ and n₂ are the principal quantum numbers.

Q8. For the first line of the Lyman series of hydrogen, calculate the wavelength λ₁ using hc/λ₁ = Rhc[1/n₁² − 1/n₂²].

1. hc/λ₁ = Rhc[1/n₁² − 1/n₂²]
2. hc/λ₁ = ZRhc[1/n₁² − 1/n₂²]
3. hc/λ₁ = R[1/n₁² − 1/n₂²]
4. hc/λ₁ = ZR[1/n₁² − 1/n₂²]

Answer: hc/λ₁ = Rhc[1/n₁² − 1/n₂²]

The formula hc/λ₁ = Rhc[1/n₁² − 1/n₂²] is correct for the Lyman series of hydrogen, where R is the Rydberg constant, and n₁ and n₂ are principal quantum numbers. This matches option A.

Q9. Which of the following is positively charged?

1. α-particle
2. β-particle
3. γ-rays
4. X-rays

Answer: α-particle

An α-particle consists of two protons and two neutrons, making it positively charged. The other options are either negatively charged (β-particle) or neutral (γ-rays and X-rays).

Q10. In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of:

1. excitation of electrons in the atoms
2. collision between the atoms of the gas
3. collisions between the charged particles emitted from the cathode and the atoms of the gas
4. collision between different electrons of the atoms of the gas

Answer: collisions between the charged particles emitted from the cathode and the atoms of the gas

The coloured glow in the discharge tube is due to collisions between the charged particles (electrons) emitted from the cathode and the atoms of the gas, which excite the gas atoms. When these excited atoms return to their ground state, they emit light of specific wavelengths, producing the glow.

Q11. The 21 cm radio wave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly

1. 10^(-17) J
2. 1 J
3. 7 × 10^(-15) J
4. 10^(-24) J

Answer: 10^(-17) J

The energy of the emitted wave can be calculated using the formula E = hν, where h is Planck's constant and ν is the frequency. For the 21 cm wavelength, the energy is approximately 10^(-17) J.

Q12. Doubly ionised helium atoms and hydrogen ions are accelerated from rest through the same potential drop. The ratio of the final velocities of the helium and the hydrogen ion is

1. 1/2
2. 2
3. 1/√2
4. √2

Answer: 1/√2

The kinetic energy gained by a charged particle accelerated through a potential difference is equal to the charge multiplied by the potential difference. Since the charge of the helium ion is twice that of the hydrogen ion, but the mass of the helium ion is four times that of the hydrogen ion, the velocity ratio is determined by the square root of the charge-to-mass ratio. This gives a ratio of 1/√2.

Q13. Energy levels A, B, C of a certain atom correspond to increasing values of energy i.e., E₄ < E₃ < E₂. λ₁, λ₂, λ₃ are the wavelengths of radiation corresponding to the transitions C to B, B to A and C to A respectively, which of the following relation is correct?

1. λ₃ = λ₁ + λ₂
2. λ₃ = λ₁λ₂ / (λ₁ + λ₂)
3. λ₁ + λ₂ + λ₃ = 0
4. λ₃² = λ₁² + λ₂²

Answer: λ₃ = λ₁λ₂ / (λ₁ + λ₂)

The energy difference for the transition C to A is the sum of the energy differences for transitions C to B and B to A. Using the relation E = hc/λ, the reciprocal of the wavelength for C to A is the sum of the reciprocals of the wavelengths for C to B and B to A. This leads to the formula λ₃ = λ₁λ₂ / (λ₁ + λ₂).

Q14. In a Rutherford scattering experiment when a projectile of charge Z₁ and mass M₁ approaches a target nucleus of charge Z₂ and mass M₂, the distance of closest approach is r₀. The energy of the projectile is

1. directly proportional to Z₁ Z₂
2. inversely proportional to Z₁
3. directly proportional to mass M₁
4. directly proportional to M₁ × M₂

Answer: directly proportional to Z₁ Z₂

The distance of closest approach in Rutherford scattering is determined by the electrostatic potential energy, which is proportional to the product of the charges Z₁ and Z₂. Hence, the energy of the projectile is directly proportional to Z₁Z₂.

Q15. In Rutherford scattering experiment, what will be the correct angle for α-scattering for an impact parameter, b = 0?

1. 90°
2. 270°
3. 0°
4. 180°

Answer: 180°

When the impact parameter b = 0, the alpha particle moves directly towards the nucleus and experiences maximum deflection due to the strong Coulomb repulsion, resulting in a scattering angle of 180°.

Q16. What is the radius of iodine atom (At. no. 53, mass no. 126)?

1. 2.5 × 10⁻¹¹ m
2. 2.5 × 10⁻⁹ m
3. 7 × 10⁻⁹ m
4. 7 × 10⁻⁶ m

Answer: 2.5 × 10⁻⁹ m

The radius of an atom is typically on the order of 10⁻¹⁰ m. For iodine, the radius is approximately 2.5 × 10⁻⁹ m, which corresponds to option B.

Q17. For which one of the following, Bohr model is not valid?

1. Singly ionised helium atom (He⁺)
2. Deuteron atom
3. Singly ionised neon atom (Ne⁺)
4. Hydrogen atom

Answer: Singly ionised neon atom (Ne⁺)

The Bohr model is valid for single-electron systems like hydrogen, He⁺, and similar ions. However, Ne⁺ has multiple electrons, making the Bohr model inapplicable due to electron-electron interactions.

Q18. The total energy of an electron in an atom in an orbit is −3.4 eV. Its kinetic and potential energies are, respectively:

1. −3.4 eV, −3.4 eV
2. −3.4 eV, −6.8 eV
3. 3.4 eV, −6.8 eV
4. 3.4 eV, 3.4 eV

Answer: 3.4 eV, −6.8 eV

In an atom, the total energy (E) of an electron is the sum of its kinetic energy (K) and potential energy (U). The kinetic energy is equal to the magnitude of the total energy, K = +3.4 eV, and the potential energy is twice the total energy but negative, U = −6.8 eV.

Q19. Who indirectly determined the mass of the electron by measuring the charge of the electron?

1. Millikan
2. Rutherford
3. Einstein
4. Thomson

Answer: Millikan

Millikan determined the charge of the electron through his oil drop experiment, which, when combined with Thomson's charge-to-mass ratio, allowed the mass of the electron to be calculated indirectly.

Q20. The radius of the first permitted Bohr orbit, for the electron, in a hydrogen atom equals 0.51 Å and its ground state energy equals −13.6 eV. If the electron in the hydrogen atom is replaced by muon (μ⁻) [charge same as electron and mass 207 mₑ], the first Bohr radius and ground state energy will be:

1. 2.56 × 10⁻¹³ m, −13.6 eV
2. 0.53 × 10⁻¹³ m, −3.6 eV
3. 25.6 × 10⁻¹³ m, −2.8 eV
4. 2.56 × 10⁻¹³ m, −2.8 eV

Answer: 2.56 × 10⁻¹³ m, −2.8 eV

The Bohr radius is inversely proportional to the mass of the orbiting particle. Since the muon is 207 times heavier than the electron, the first Bohr radius will be reduced by a factor of 207, giving approximately 2.56 × 10⁻¹³ m. The energy is inversely proportional to the radius, so the ground state energy will increase by a factor of 207, resulting in −2.8 eV.