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प्रश्न
How will the de-Broglie wavelength associated with an electron be affected when the velocity of the electron decreases? Justify your answer.
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उत्तर
According to the de-Broglie wavelength formula, the wavelength is inversely proportional to electron momentum. The product of mass and velocity is momentum. As a result, as the electron's velocity declines, so does its momentum. As a result, the de-Broglie wavelength of the electron will grow.
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संबंधित प्रश्न
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Describe briefly how the Davisson-Germer experiment demonstrated the wave nature of electrons.
Sodium and copper have work function 2.3 eV and 4.5 eV respectively. Then, the ratio of the wavelengths is nearest to ______.
Two particles A1 sand A2 of masses m1, m2 (m1 > m2) have the same de Broglie wavelength. Then ______.
- their momenta are the same.
- their energies are the same.
- energy of A1 is less than the energy of A2.
- energy of A1 is more than the energy of A2.
The de Broglie wavelength of a photon is twice the de Broglie wavelength of an electron. The speed of the electron is `v_e = c/100`. Then ______.
- `E_e/E_p = 10^-4`
- `E_e/E_p = 10^-2`
- `p_e/(m_ec) = 10^-2`
- `p_e/(m_ec) = 10^-4`
A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values λ1, λ2 with λ1 > λ2. Which of the following statement are true?
- The particle could be moving in a circular orbit with origin as centre.
- The particle could be moving in an elliptic orbit with origin as its focus.
- When the de Broglie wavelength is λ1, the particle is nearer the origin than when its value is λ2.
- When the de Broglie wavelength is λ2, the particle is nearer the origin than when its value is λ1.
Assuming an electron is confined to a 1 nm wide region, find the uncertainty in momentum using Heisenberg Uncertainty principle (∆x∆p ≃ h). You can assume the uncertainty in position ∆x as 1 nm. Assuming p ≃ ∆p, find the energy of the electron in electron volts.
An alpha particle is accelerated through a potential difference of 100 V. Calculate:
- The speed acquired by the alpha particle, and
- The de-Broglie wavelength is associated with it.
(Take mass of alpha particle = 6.4 × 10−27 kg)
The ratio of wavelengths of proton and deuteron accelerated by potential Vp and Vd is 1 : `sqrt2`. Then, the ratio of Vp to Vd will be ______.
The graph which shows the variation of `(1/lambda^2)` and its kinetic energy, E is (where λ is de Broglie wavelength of a free particle):
