#work function

PHYSICS

The threshold wavelength of a metal is 550 nm . What is its work function in joules? (Take h = 6.63 × 10⁻³⁴ J s, c

Given: The threshold wavelength of a metal is 550 nm . What is its work function in joules? (Take h = 6.63 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s ) These values define the system as per NCERT data. Formula: v_0 = c/lambda_0 = frac3 × 10⁸⁵⁵⁰ × 10⁻⁹ approx 5.455 × 10¹⁴ Hz. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: phi_0 = h v_0 = 6.63 × 10⁻³⁴ × 5.455 × 10¹⁴ approx 3.617 × 10⁻¹⁹ J . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Oscillations and Waves, Topic: SHM, wave motion and superposition.

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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

A metal has a work function of 2.5 eV. What is the threshold wavelength? (h = 6.626 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m s⁻¹, 1

λ₀ = hc/W₀ ≈ 497 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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Shapes of Orbitals - s p d f and Nodes and Radial Probability

A photon of wavelength 300 nm strikes a metal surface, and the ejected electron has a kinetic energy of 2.0 × 10⁻¹⁹ J. W

E = hc/λ = 6.626×10⁻¹⁹ J. W₀ = E - KE = 4.626×10⁻¹⁹ 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

The work function of a metal is 2.5 eV. What is the maximum wavelength of light that can eject electrons from this metal

W₀ = 4.0×10⁻¹⁹ J. λmax = hc/W₀ ≈ 4.97×10⁻⁷ m = 497 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 Numbers - Principal Azimuthal Magnetic Spin

A metal has a work function of 3.2 eV. What is the kinetic energy of an electron ejected by light of wavelength 350 nm?

E ≈ 3.55 eV. KE ≈ 0.35 eV ≈ 5.59×10⁻²⁰ J.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Numbers - Principal Azimuthal Magnetic Spin

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Quantum Numbers - Principal Azimuthal Magnetic Spin

The threshold frequency for a metal is 6.0 × 10¹⁴ Hz. What is the work function in joules? (h = 6.626 × 10⁻³⁴ J s)

W₀ = hν₀ = 3.9756×10⁻¹⁹ J.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Numbers - Principal Azimuthal Magnetic Spin

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Quantum Numbers - Principal Azimuthal Magnetic Spin

A photon of energy 5.0 × 10⁻¹⁹ J strikes a metal with a work function of 3.5 × 10⁻¹⁹ J. What is the velocity of the ejec

KE = 1.5×10⁻¹⁹ J. v = √(2KE/m) ≈ 5.74×10⁵ m s⁻¹.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Numbers - Principal Azimuthal Magnetic Spin

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Quantum Numbers - Principal Azimuthal Magnetic Spin

A photon of frequency 7.5 × 10¹⁴ Hz ejects an electron from a metal with a kinetic energy of 1.9878 × 10⁻¹⁹ J. What is t

E = hν ≈ 4.970×10⁻¹⁹ J. W₀ = E - KE ≈ 2.982×10⁻¹⁹ J.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Quantum Numbers - Principal Azimuthal Magnetic Spin

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Quantum Mechanical Model - de Broglie Heisenberg and Schrodinger

A photon of wavelength 250 nm strikes a metal surface. The ejected electron has a de Broglie wavelength of 1.0 × 10⁻⁹ m.

E = hc/λ ≈ 7.951×10⁻¹⁹ J. KE = h²/(2mλ²) ≈ 2.412×10⁻¹⁹ J. W₀ = E - KE ≈ 5.539×10⁻¹⁹ J.

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

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Electronic Configuration - Aufbau Pauli Hund and Stability

A photon of wavelength 200 nm ejects an electron from a metal with a velocity of 6.62 × 10⁵ m s⁻¹. What is the work func

E ≈ 9.939×10⁻¹⁹ J. KE ≈ 1.994×10⁻¹⁹ J. W₀ ≈ 7.945×10⁻¹⁹ J.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability

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Electronic Configuration - Aufbau Pauli Hund and Stability

The kinetic energy of a photoelectron is 1.5 × 10⁻¹⁹ J when light of wavelength 400 nm is used. What is the work functio

E = hc/λ ≈ 4.970×10⁻¹⁹ J. W₀ = E - KE ≈ 3.470×10⁻¹⁹ J.

Ref: NCERT Class 11 Chemistry > Chapter 2: Structure of Atom > Topic: Electronic Configuration - Aufbau Pauli Hund and Stability

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