#physics exam

Third Law of Thermodynamics and Applications

For an ideal gas (γ = 1.33) compressed adiabatically from 18 L to 6 L at 300 K, what is the final temperature?

T₂ = T₁ (V₁/V₂)γ⁻¹ = 300 × (18/6)⁰.³³ = 300 × 3⁰.³³ = 300 × 1.44 = 432 K.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Third Law of Thermodynamics and Applications

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Gibbs Energy and Gibbs Energy Change and Equilibrium

In an adiabatic irreversible expansion of an ideal gas against a constant external pressure, the work done is:

For irreversible expansion against constant pext, w = -pext Δ V, where Δ V is the volume change.

Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Gibbs Energy and Gibbs Energy Change and Equilibrium

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Shapes of Orbitals - s p d f and Nodes and Radial Probability

The energy required to eject an electron from a metal is 4.5 × 10⁻¹⁹ J. If light of frequency 8.0 × 10¹⁴ Hz is used, wha

E ≈ 5.301×10⁻¹⁹ J. KE ≈ 8.008×10⁻²⁰ J.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Shapes of Orbitals - s p d f and Nodes and Radial Probability

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Shapes of Orbitals - s p d f and Nodes and Radial Probability

Which series of spectral lines in a hydrogen atom lies in the visible region?

The Balmer series lies in the visible region.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Shapes of Orbitals - s p d f and Nodes and Radial Probability

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Shapes of Orbitals - s p d f and Nodes and Radial Probability

The ionization energy of Li²⁺ is 122.4 eV. What is the wavelength of light required to excite an electron from n = 1 to

ΔE = 108.8 eV. λ ≈ 11.42 nm.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Shapes of Orbitals - s p d f and Nodes and Radial Probability

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Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

The velocity of an electron in the second orbit of a hydrogen atom is 1.095 × 10⁶ m s⁻¹. What is its de Broglie waveleng

λ = h/(mv) = 6.65×10⁻¹⁰ m.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

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Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

What is the wavelength of the second line in the Balmer series of a hydrogen atom? (RH = 1.097 × 10⁷ m⁻¹)

n=4 to n=2. λ = 486.1 nm.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

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Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

An electron and an alpha particle have the same de Broglie wavelength. What is the ratio of their velocities? (mₑ = 9.1

vₑ/v_α = m_α/mₑ ≈ 7297.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

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Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

How many electrons in an atom can have the quantum numbers n = 5 and l = 2?

For l=2 (d subshell), orbitals = 5. Electrons = 5×2 = 10.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

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Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

A proton and an alpha particle have the same de Broglie wavelength. What is the ratio of their velocities? (mₚ = 1.67 ×

mₚ vₚ = m_α v_α. vₚ/v_α = m_α/mₚ ≈ 4.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

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Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

A metal has a threshold wavelength of 500 nm. What is the stopping potential when irradiated with light of wavelength 40

W₀ = hc/λ₀ ≈ 3.976×10⁻¹⁹ J. E = hc/λ ≈ 4.970×10⁻¹⁹ J. KE ≈ 9.94×10⁻²⁰ J. Vₛ = KE/e ≈ 0.621 V.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

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Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

A proton has a de Broglie wavelength of 1.5 × 10⁻¹² m. What is its kinetic energy? (h = 6.626 × 10⁻³⁴ J s, mₚ = 1.67 × 1

v = h/(mλ) = 2.645×10⁵ m s⁻¹. KE = ½mv² = 5.837×10⁻¹⁷ J.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

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