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प्रश्न
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.
पर्याय
a and c
a and d
c and d
a and b
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उत्तर
a and c
Explanation:
We know that de-Broglie wavelength `λ = h/(mv)`
Where, mv = p (momentum) of the particle
⇒ But we can express wavelength `λ = h/p` ⇒ `p = h/λ`
Hence, `p oo 1/λ ⇒ p_1/p_2 = λ_2/λ_1`
But particles have the same de-Broglie wavelength `(λ_1 = λ_2) = λ`
Then `p_1/p_2 = λ/λ` = 1 ⇒ `p_1 = p_2`
Thus, their momenta are the same.
Also, `E = 1/2 mv^2 = 1/2 mv^2 m/m`
= `1/2 (m^2v^2)/m = 1/2 p^2/m`
As p is constant, `E oo 1/m`
∴ `E_1/E_2 = m_2/m_1 < 1 ⇒ E_1 < E_2`
Important points:
Some important characteristics of Matter Waves:
- Matter wave represents the probability of finding a particle in space.
- Matter waves are not electromagnetic in nature.
- de-Broglie or matter wave is independent of the charge on the material particle. It means matter-wave of de-Broglie wave is associated with every moving particle (whether charged or uncharged).
- Practical observation of matter waves is possible only when the de-Broglie wavelength is of the order of the size of the particles.
- Electron microscope works on the basis of de-Broglie waves.
- The phase velocity of the matter waves can be greater than the speed of light.
- Matter waves can propagate in a vacuum, hence they are not mechanical waves.
- The number of de-Broglie waves associated with nth orbital electron is n.
- Only those circular orbits around the nucleus are stable whose circumference is an integral multiple of de-Broglie wavelength associated with the orbital electron.
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