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#momentum

26 public questions tagged with this topic.

Photon Energy, Momentum, Power and X-rays

A photon has a wavelength of \( 200 \, \text{nm} \). What is its momentum? (Take \( h = 6.63 \times 10^{-34} \, \text{J

**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. p = (h/λ) = (6.63 × 10⁻³⁴/200 × 10⁻⁹) = 3.315 × 10⁻²⁷ kg m/s . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h

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

In the context of matter waves, what property of a moving particle is inversely proportional to its de Broglie wavelengt

**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 is λ = (h/p) , where p (momentum) is inversely proportional to λ . 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

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

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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,

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

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PHYSICS

The momentum p depends on mass m and velocity v as p = k m^a v^b . Given [p] = [M L T^{-1], find a and b .

Given: The momentum p depends on mass m and velocity v as p = k m^a v^b . Given [p] = [M L T^{-1], find a and b . These values define the system as per NCERT data. Formula: [m] = [M], [v] = [L T^{-1]. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: [M L T^{-1] = [M]^a [L T^{-1]^b = [M^a L^b T^{-b] . Equate: a = 1, b = 1, -b = -1 Rightarrow b = 1 . a = 1, b = 1 . 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: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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Work, Energy and Laws of Motion - Mixed Practice

A 8kg mass at 12m/s collides elastically with an identical stationary mass. What is the speed of the second mass after c

For elastic collision, momentum and kinetic energy conserved: m1v1+m2v2=m1v1'+m2v2' and ½m1v1²+... per NCERT. Solving gives velocities matching 12 m/s. Inelastic loses energy. So option C is correct.

Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Work, Energy and Laws of Motion - Mixed Practice

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Straight Line Motion - Practice Set

A 10kg mass at 8m/s collides inelastically with a stationary 5kg mass. What is the final speed?

Momentum conservation: 10×8=(10+5)vf⇒80=15vf⇒vf=8015≈5.33m/s. 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.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0

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Straight Line Motion - Practice Set

A 0.2kg ball moving at 30m/s at 45∘ to a wall rebounds at the same speed and angle. What is the impulse perpendicular to

Perpendicular component (x-axis) changes direction upon rebound. 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 6 N s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0

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