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प्रश्न
A particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is proportional to ______.
पर्याय
`H`
`H^(1/2)`
`H^0`
`H^(-1/2)`
उत्तर
A particle is dropped from a height H. The de Broglie wavelength of the particle as a function of height is proportional to `underline(H^(-1/2))`.
Explanation:
According to de-Broglie a moving material particle sometimes acts as a wave and sometimes as a particle.
The wave associated with a moving particle is called matter wave or de-Broglie wave and it propagates in the form of wave packets with the group velocity. According to de-Broglie theory, the wavelength of de-Broglie wave is given by `H = v = sqrt(2gH)`
We know that de-Broglie wavelength `λ = h/p`
`λ = h/(mv) = h/(msqrt(2gH)`
h, m and g are constant
∴ `h/(msqrt(2g)` is constant ⇒ `λ oo 1/sqrt(H)` ⇒ `λ oo H^(-1/2)`
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संबंधित प्रश्न
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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.
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- 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.
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The De-Broglie wavelength of electron in the third Bohr orbit of hydrogen is ______ × 10-11 m (given radius of first Bohr orbit is 5.3 × 10-11 m):
The equation λ = `1.227/"x"` nm can be used to find the de Brogli wavelength of an electron. In this equation x stands for:
Where,
m = mass of electron
P = momentum of electron
K = Kinetic energy of electron
V = Accelerating potential in volts for electron
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