#particle speed

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 -

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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Photon Energy, Momentum, Power and X-rays

A particle of mass \( 1.5 \times 10^{-30} \, \text{kg} \) has the same momentum as a photon of wavelength \( 500 \, \tex

**Photon power** light source emits 1.0×10¹⁶ photons/s power 4.0 mW, energy per photon E= P/N =4×10⁻³/10¹⁶=4×10⁻¹⁹ J, wavelength λ= h c/E=6.63×10⁻³⁴×3×10⁸/4×10⁻¹⁹=497 nm, for 6×10¹⁵ photons/s 3.0 mW E=5×10⁻¹⁹ J, for beam 2.5×10¹⁵ photons/s 4×10⁻¹⁹ J each power=10⁻³ W=1 mW, illustrating P = N h f. Photon momentum p = (h/λ) = (6.63 × 10⁻³⁴/500 × 10⁻⁹) = 1.326 × 10⁻²⁷ kg m/s . v = (p/m) = (1.326 × 10⁻²⁷/1.5 × 10⁻³⁰) = 8.84 × 10² = 884 m/s . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ

Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays

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Photon Energy, Momentum, Power and X-rays

A particle of mass \( 1.0 \times 10^{-28} \, \text{kg} \) has the same momentum as a photon of wavelength \( 600 \, \tex

**Photon energy** E = h f = h c/λ, momentum p = h/λ = E/c, power P = N E/t, N number of photons per second, wavelength λ = h c/E, for photon momentum 1.5×10⁻²⁷ kg·m/s frequency f = p c/h? Actually E = p c, f = E/h = p c/h =1.5×10⁻²⁷×3×10⁸/6.63×10⁻³⁴=6.79×10¹⁴ Hz, energy E= h f =4.5×10⁻¹⁹ J, for 5.0×10⁻¹⁹ J frequency f=E/h=7.54×10¹⁴ Hz. Photon momentum p = (h/λ) = (6.63 × 10⁻³⁴/600 × 10⁻⁹) = 1.105 × 10⁻²⁷ kg m/s . For particle, p = m v ⇒ v = (p/m) = (1.105 × 10⁻²⁷/1.0 × 10⁻²⁸) = 1.105 × 10¹ =

Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays

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De Broglie Wavelength and Matter Waves

A particle of mass \( 2.5 \times 10^{-29} \, \text{kg} \) has a de Broglie wavelength of \( 0.5 \, \text{nm} \). What is

**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. p = (h/λ) = (6.63 × 10⁻³⁴/0.5 × 10⁻⁹) = 1.326 × 10⁻²⁴ kg m/s . v = (p/m) = (1.326 × 10⁻²⁴/2.5 × 10⁻²⁹) = 5.304 × 10⁴ m/s . 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 5.304 × 10⁴ m/s follows, reflecting photoelectric and de Broglie relations.

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

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Motion Parameters and Calculations

A particle moves with an initial velocity of 4i^m/s and an acceleration of (2i^−3j^)m/s2. What is its speed after 3s?

Velocity v=v0+at. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 10 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion Parameters and Calculations

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Acceleration and Velocity Analysis

A particle starts with velocity 7j^m/s and accelerates at (3i^−2j^)m/s2. What is its speed after 2s?

Velocity v=v0+at. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 5 m/s as the result, so option A is correct.