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When would the wavelength associated with ,as electron become equal to the wavelength associated with proton?? plz help friends... waiting for your answer
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When would the wavelength associated with ,as electron become equal to...
**When would the wavelength associated with an electron become equal to the wavelength associated with a proton?**

To understand when the wavelength associated with an electron becomes equal to the wavelength associated with a proton, we need to consider the de Broglie wavelength and the factors that affect it.

**De Broglie Wavelength:**
According to Louis de Broglie's hypothesis, every particle with momentum exhibits wave-like properties. The de Broglie wavelength (λ) is the wavelength associated with a particle and is given by the equation:

λ = h / p

Where:
λ - de Broglie wavelength
h - Planck's constant (6.626 x 10^-34 J·s)
p - momentum of the particle

The momentum (p) of a particle is given by the equation:

p = mv

Where:
m - mass of the particle
v - velocity of the particle

**Electron vs. Proton:**
Electrons and protons are subatomic particles that have different masses and charges. The mass of an electron (me) is approximately 9.11 x 10^-31 kg, while the mass of a proton (mp) is approximately 1.67 x 10^-27 kg. Additionally, electrons have a negative charge (-1) while protons have a positive charge (+1).

**Factors affecting de Broglie Wavelength:**
The de Broglie wavelength depends on the momentum of the particle, which in turn depends on the mass and velocity. Therefore, the factors that affect the de Broglie wavelength include:

1. Mass of the particle: As the mass increases, the momentum decreases, resulting in a longer de Broglie wavelength.
2. Velocity of the particle: As the velocity increases, the momentum increases, resulting in a shorter de Broglie wavelength.

**Equalizing the Wavelengths:**
To find when the de Broglie wavelength associated with an electron becomes equal to the de Broglie wavelength associated with a proton, we need to equate their wavelengths:

λe (electron) = λp (proton)

Using the equations for the de Broglie wavelength and momentum, we can rearrange and solve for the velocities of the electron and proton:

λe = h / pe
λp = h / pp

Since the masses of the electron and proton are different, their momenta will also be different for the same de Broglie wavelength. Therefore, the velocities of the electron and proton will be different as well.

In conclusion, the wavelengths associated with an electron and a proton will never be equal as their masses and charges differ. The de Broglie wavelength is dependent on the mass and velocity of the particle, and since electrons and protons have different masses and charges, their de Broglie wavelengths will always be different.
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When would the wavelength associated with ,as electron become equal to...
ngthλ = hmvAs its given that the conditionis wavelength of electron and proton are equal,,, Let us equate likehmv = hMVWhere m-mass of electronM -mass of protonv -velocity of electgronV -Velocity of protonv = MVm =1.6726 × 10-27 × V9.1 × 10-31 = 1837VWhen the velocity of electron becomes 1837 times the velocity of protons, the wavelengths are equal , it would
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The French physicist Louis de-Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength of a material particle, its linear momentum p and planck constant h.The de Broglie relation implies that the wavelength of a particle should decreases as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they,(i) have lower velocities(ii) have no electrical and magnetic fields and(iii) are not emitted by the particle under consideration.The experimental confirmation of the deBroglie relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave, as proposed by deBroglie.Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron. He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also ture. This is summed up in what we now call the Heisenberg uncertainty principle : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of undertainty in the position, x and the uncertainity in the momentum (mv) must be greater than or equal to h/4. i.e.Q. The transition, so that the de - Broglie wavelength of electron becomes 3 times of its initial value in He+ ion will be

When would the wavelength associated with ,as electron become equal to the wavelength associated with proton?? plz help friends... waiting for your answer
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