Matter waves, de Broglie relation
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See all →- De-Broglie wavelength of an electron orbiting in the n = 2 state of hydrogen atom is close to (Given Bohr radius = 0.052 nm) Answer: 2. 0.67 nm Key idea: Bohr's quantization 2π r_n=nλ (electron wave fits the orbit). Radius of the n=2 orbit: r₂=n²a₀=4×0.052=0.208 nm... 1
- A photon and an electron (mass m) have the same energy E. The ratio ≤ft(λ_photon/λ_electron) of their de Broglie wavelengths is: (c is the speed of light) Answer: 3. c√(2m/E) Key idea: λ=h/p , with different momentum-energy relations for photon and electron. Photon: E=pc⇒λ_ph=hc/E . Electron: E... 1
- The graph which shows the variation of ≤ft(1/λ^2) and its kinetic energy, E is (where λ is de Broglie wavelength of a free particle) : Answer: 1. graph (1) Key idea: For a free particle E=p²/2m and λ=h/p . So E=h²/2mλ² , which gives frac1λ²=2m/h²E . frac1λ² is directly propo... 1
- The graph which shows the variation of the de Broglie wavelength (λ) of a particle and its associated momentum (p) is : Answer: 1. (graph) de Broglie wavelength is λ=h/p , so λ is inversely proportional to p . The graph of λ against p is therefore a rectangula... 1
- An electromagnetic wave of wavelength 'λ' is incident on a photosensitive surface of negligible work function. If 'm' mass is of photoelectron emitted from the surface has de-Broglie wavelength λ_d, then : Answer: 3. λ = (2mc/h)λ_d² Key idea: with negligible work function, the photoelectron's kinetic energy equals the photon energy. hc/λ=p²/2m⇒... 1
- An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227×10^-2 nm, the potential difference is : Answer: 4. 10⁴ V de Broglie wavelength of an electron accelerated through V volts: λ=1.227/sqrt V nm. Given λ=1.227×10⁻² nm: sqrt V=frac1.22... 1