#quantum mechanics

De Broglie Wavelength and Matter Waves

The de Broglie wavelength of a particle of mass \( 2.0 \times 10^{-30} \, \text{kg} \) moving at \( 1.5 \times 10^6 \, \

**Matter waves** wave-like properties of moving particles, macroscopic objects not exhibit measurable wave-like because λ = h/(m v) extremely small for large m, e.g., 1 kg at 1 m/s λ=6.6×10⁻³⁴ m undetectable, for electron 9.11×10⁻³¹ kg at 2×10⁶ m/s λ=0.36 nm measurable via diffraction, de Broglie hypothesis λ = h/p proposed by de Broglie 1924, Davisson-Germer experiment confirmed. p = m v = 2.0 × 10⁻³⁰ × 1.5 × 10⁶ = 3.0 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/3.0 × 10⁻²⁴) = 2.21 × 10⁻¹⁰ m = 0.221 nm . Applying E = h f = h c/λ, p =

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De Broglie Wavelength and Matter Waves

The de Broglie wavelength of an electron is \( 0.2 \, \text{nm} \). What is its speed? (Take \( h = 6.63 \times 10^{-34}

**Matter waves** wave-like properties of moving particles, macroscopic objects not exhibit measurable wave-like because λ = h/(m v) extremely small for large m, e.g., 1 kg at 1 m/s λ=6.6×10⁻³⁴ m undetectable, for electron 9.11×10⁻³¹ kg at 2×10⁶ m/s λ=0.36 nm measurable via diffraction, de Broglie hypothesis λ = h/p proposed by de Broglie 1924, Davisson-Germer experiment confirmed. p = (h/λ) = (6.63 × 10⁻³⁴/0.2 × 10⁻⁹) = 3.315 × 10⁻²⁴ kg m/s . v = (p/m) = (3.315 × 10⁻²⁴/9.11 × 10⁻³¹) ≈ 3.64 × 10⁶ m/s . Applying E = h f = h c/λ, p = h/λ, K_max = h f -

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De Broglie Wavelength and Matter Waves

The de Broglie wavelength of a particle of mass \( 5.0 \times 10^{-30} \, \text{kg} \) moving at \( 2.0 \times 10^5 \, \

**De Broglie wavelength calculations** for electron moving at 2×10⁶ m/s λ= h/(m v)=6.63×10⁻³⁴/(9.11×10⁻³¹×2×10⁶)=3.64×10⁻¹⁰ m=0.364 nm, for mass 5×10⁻³⁰ kg v=2×10⁵ m/s λ=6.63×10⁻³⁴/(1×10⁻²⁴)=6.63×10⁻¹⁰ m, showing λ decreases with increasing m v. Momentum p = m v = 5.0 × 10⁻³⁰ × 2.0 × 10⁵ = 1.0 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/1.0 × 10⁻²⁴) = 6.63 × 10⁻¹⁰ m = 0.663 nm . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ = h/√(2 m e V) and f_max = e V/h, result 0.663

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De Broglie Wavelength and Matter Waves

An electron moves with a speed of \( 3.0 \times 10^6 \, \text{m/s} \). What is its de Broglie wavelength? (Take \( h = 6

**Matter waves** wave-like properties of moving particles, macroscopic objects not exhibit measurable wave-like because λ = h/(m v) extremely small for large m, e.g., 1 kg at 1 m/s λ=6.6×10⁻³⁴ m undetectable, for electron 9.11×10⁻³¹ kg at 2×10⁶ m/s λ=0.36 nm measurable via diffraction, de Broglie hypothesis λ = h/p proposed by de Broglie 1924, Davisson-Germer experiment confirmed. Momentum p = m v = 9.11 × 10⁻³¹ × 3.0 × 10⁶ = 2.733 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/2.733 × 10⁻²⁴) ≈ 2.425 × 10⁻¹⁰ m = 0.2425 nm . Applying E = h f = h c/λ, p

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De Broglie Wavelength and Matter Waves

Why do macroscopic objects not exhibit measurable wave-like properties according to de Broglie’s hypothesis?

**De Broglie wavelength** λ = h/p = h/(m v) = h/√(2 m e V) for electron accelerated through V, for V=100 V λ= h/√(2 m e×100)=1.227/√V nm=0.1227 nm, for 50 V 0.173 nm, for particle mass 3.0×10⁻³⁰ kg v=10⁶ m/s λ=6.63×10⁻³⁴/(3×10⁻³⁰×10⁶)=2.21×10⁻¹⁰ m=0.221 nm, inversely proportional to momentum, property inversely proportional to de Broglie wavelength is momentum. The de Broglie wavelength λ = (h/p) is extremely small for macroscopic objects due to their large momentum, making wave effects negligible. Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ = h/√(2

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De Broglie Wavelength and Matter Waves

An electron is accelerated through \( 50 \, \text{V} \). What is its de Broglie wavelength? (Take \( h = 6.63 \times 10^

**Matter waves** wave-like properties of moving particles, macroscopic objects not exhibit measurable wave-like because λ = h/(m v) extremely small for large m, e.g., 1 kg at 1 m/s λ=6.6×10⁻³⁴ m undetectable, for electron 9.11×10⁻³¹ kg at 2×10⁶ m/s λ=0.36 nm measurable via diffraction, de Broglie hypothesis λ = h/p proposed by de Broglie 1924, Davisson-Germer experiment confirmed. K = e V = 1.6 × 10⁻¹⁹ × 50 = 8.0 × 10⁻¹⁸ J . p = √(2 m K) = √(2 × 9.11 × 10⁻³¹ × 8.0 × 10⁻¹⁸) ≈ 3.816 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/3.816 × 10⁻²⁴)

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De Broglie Wavelength and Matter Waves

The de Broglie wavelength of an electron moving at \( 2.0 \times 10^6 \, \text{m/s} \) is calculated. What is its value?

**De Broglie wavelength** λ = h/p = h/(m v) = h/√(2 m e V) for electron accelerated through V, for V=100 V λ= h/√(2 m e×100)=1.227/√V nm=0.1227 nm, for 50 V 0.173 nm, for particle mass 3.0×10⁻³⁰ kg v=10⁶ m/s λ=6.63×10⁻³⁴/(3×10⁻³⁰×10⁶)=2.21×10⁻¹⁰ m=0.221 nm, inversely proportional to momentum, property inversely proportional to de Broglie wavelength is momentum. Momentum p = m v = 9.11 × 10⁻³¹ × 2.0 × 10⁶ = 1.822 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/1.822 × 10⁻²⁴) ≈ 3.64 × 10⁻¹⁰ m = 0.364 nm . Applying E = h f = h c/λ, p = h/λ,

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Hydrogen Atom Properties - Radius, Speed and Energy

An electron in a hydrogen atom has a total energy of -3.4 eV. What is its kinetic energy?

**Energy in hydrogen** total E_n = -13.6/n² eV, kinetic K = +13.6/n² eV, potential U = -27.2/n² eV, ratio K/U = -1/2, total negative indicates bound, zero at ionization, negative total means electron bound, requires energy to free. For n=4 E=-0.85 eV, U=-1.7 eV, K=0.85 eV, potential energy twice total negative. E = -K , K = -E = -(-3.4) = 3.4 eV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 3.4 eV, consistent with Bohr model and nuclear binding energy systematics.

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Hydrogen Atom Properties - Radius, Speed and Energy

In a hydrogen atom, the total energy of the electron in the ground state is -13.6 eV. What is its kinetic energy?

**Energy in hydrogen** total E_n = -13.6/n² eV, kinetic K = +13.6/n² eV, potential U = -27.2/n² eV, ratio K/U = -1/2, total negative indicates bound, zero at ionization, negative total means electron bound, requires energy to free. For n=4 E=-0.85 eV, U=-1.7 eV, K=0.85 eV, potential energy twice total negative. Total energy E = K + U , where K = (e²/8πepsilon₀ r) , U = -(e²/4πepsilon₀ r) . E = K - 2K = -K . E = -13.6 eV = -K ⇒ K = 13.6 eV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R =

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De Broglie Hypothesis and Quantization in Bohr Model

In the Bohr model, how many de Broglie wavelengths fit into the circumference of the \( n = 3 \) orbit?

**Quantization basis** de Broglie standing wave requires constructive interference, integer wavelengths in orbit, otherwise destructive, so only certain radii allowed r_n = n² a₀, a₀=0.53 Å, angular momentum L = n h/2π, de Broglie explains why orbits are stationary - electron wave closed on itself. 2π r_n = nλ . For n = 3 , number of wavelengths = 3. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields 3, consistent with Bohr model and nuclear binding energy systematics.

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Hydrogen Atom Properties - Radius, Speed and Energy

In a hydrogen atom, the total energy of an electron in the ground state is -13.6 eV. What is the magnitude of its potent

**Energy in hydrogen** total E_n = -13.6/n² eV, kinetic K = +13.6/n² eV, potential U = -27.2/n² eV, ratio K/U = -1/2, total negative indicates bound, zero at ionization, negative total means electron bound, requires energy to free. For n=4 E=-0.85 eV, U=-1.7 eV, K=0.85 eV, potential energy twice total negative. E = K + U , K = 13.6 eV , U = -2K = -27.2 eV . Magnitude = |U| = 27.2 eV . Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R = R₀ A^¹/³, BE = Δm c² and 1 u = 931.5 MeV, evaluation yields

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Atomic Models - Rutherford, Thomson and Bohr

In Bohr’s model, what prevents an electron from emitting radiant energy while revolving in a stable orbit?

**Thomson's plum pudding model** positive charge uniformly distributed in sphere with electrons embedded, positive charge spread, fails to explain large angle scattering observed. Bohr's model introduces stationary orbits with quantized angular momentum L = n h/2π, physical basis de Broglie standing wave condition 2πr = n λ, circumference fits n wavelengths, explains line spectrum. Bohr’s first postulate states that electrons in certain stable orbits (stationary states) do not emit radiant energy, contrary to classical electromagnetic theory. Using E_n = -13.6/n² eV, r_n = n² a₀, L = n h/2π, R

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