De Broglie Hypothesis

To determine the de Broglie wavelength of the electron at time , we need to consider the principles of the de Broglie wavelength and the effect of the electric field on the motion of the electron.

Initially, the de Broglie wavelength is given by:

where is Planck's constant, is the mass of the electron, and is its initial velocity.

The force on the electron due to the electric field is:

where is the charge of the electron.

The resulting acceleration is given by:

The velocity of the electron at time can be obtained using the equation of motion:

The de Broglie wavelength at time is:

Substituting the expression for into the de Broglie wavelength formula, we have:

Rearranging this in terms of , we get:

Thus, the de Broglie wavelength after time for the electron is:

This matches the correct answer:

Hence, the correct answer is the option:

To find the ratio of kinetic energies of a proton and an electron given they have the same de-Broglie wavelength, we need to follow several steps. The de-Broglie wavelength is given by the formula:

where is Planck's constant and is the momentum. Since both the proton and electron have the same , their momenta are equal:

The momentum can also be expressed as:

where is the mass and is the kinetic energy. Setting the momenta of the electron and proton equal gives:

Squaring both sides, we get:

Simplifying, we have:

From this, the ratio of kinetic energies and is:

Given that the mass of the proton is 1836 times the mass of the electron , we substitute:

Thus, the ratio of the kinetic energies of the electron to the proton is given by:

This confirms that the correct answer is 1 : 1836.

To solve this problem, we need to compare the kinetic energy of an electron and a photon given that they have the same de-Broglie wavelength.

First, let's understand the de-Broglie wavelength formula:

where is the de-Broglie wavelength, is Planck's constant, and is the momentum.

For an electron, the momentum is given by:

where is the mass of the electron and is its velocity.

For a photon, the momentum is:

where is the energy of the photon and is the speed of light. For a photon, energy and is the frequency of the photon.

Given that the velocity of the electron is 25% of the speed of light, we have:

The kinetic energy (K.E.) of the electron is:

K.E._e = \frac{1}{2} m v^2\)

Since the de-Broglie wavelength of the electron and the photon is the same, their momenta are equal:

The kinetic energy of the photon (considering relativistic energy) is purely its energy due to its frequency, which relates to its momentum:

K.E._p = p_p c\)

The ratio of kinetic energies is given by:

Replacing with from the equality of momenta:

After simplifying:

Therefore, the ratio of the kinetic energy of the electron to the kinetic energy of the photon is . Thus, the correct answer is:

.

To compare the de-Broglie wavelengths of particles with the same energies, we need to understand the de-Broglie wavelength formula and how it is affected by the mass and energy of the particles.

The de-Broglie wavelength is given by the formula:

where:

• is the Planck's constant.
• is the momentum of the particle.

For a particle with kinetic energy , the momentum can be expressed as:

Substituting this in the de-Broglie equation, we get:

From this equation, we see that for particles with the same energy , the wavelength is inversely proportional to the square root of their masses:

Now, let's compare the masses of the given particles:

• Proton ( ): Approximately kg
• Electron ( ): Approximately kg
• Alpha particle ( ): Consists of 2 protons and 2 neutrons, approximately kg = kg

Based on these masses:

• The electron is the lightest, hence it has the longest wavelength.
• The proton is heavier than the electron, but lighter than the alpha particle, so it has an intermediate wavelength.
• The alpha particle is the heaviest, hence it has the shortest wavelength.

Therefore, the correct order of the de-Broglie wavelengths is:

Thus, the correct answer is:

To solve this problem, we need to understand the concept of de Broglie wavelength and how it relates to the kinetic energy of particles like protons and electrons.

The de Broglie wavelength ( ) of a particle is given by the equation:

where is Planck's constant and is the momentum of the particle.

The momentum ( ) can also be related to kinetic energy ( ) using the relation:

where is the mass of the particle.

Given that the de Broglie wavelengths of a proton and an electron are the same, we can set up the equality:

From this, we can cancel out Planck's constant ( ) and rearrange to find:

Squaring both sides gives:

Solving for , we have:

Since the mass of the proton ( ) is much greater than the mass of the electron ( ), it follows that:

This shows that the kinetic energy of the proton is less than the kinetic energy of the electron when both have the same de Broglie wavelength.

Thus, the correct answer is:

The given problem requires us to find the dimensional formula for Planck's constant . The de-Broglie wavelength formula given is:

To find the dimensional formula of , we will utilize the given equation and analyze the dimensions involved. The de-Broglie wavelength formula can be rearranged as:

Let's express the dimensions of each component:

• Wavelength : This is a length, and its dimensional formula is .
• Mass : Its dimensional formula is .
• Energy : Energy has a dimensional formula of .

Using these, we calculate the dimensions for :

Simplifying, we get:

Substituting back into the formula for :

Therefore, the dimensional formula for Planck's constant is:

[ML²T^-1]

This matches the provided correct answer option, which is:

To conclude, the correct dimensional formula for Planck's constant is indeed , and the selected option is correct. The other options do not match this dimensional analysis.

The Correct Answer is :

To determine the correct graphical representation of the relationship between de Broglie wavelength and kinetic energy for a particle of constant mass, we start with the de Broglie wavelength formula: , where is Planck's constant and is momentum. The momentum of a particle is given by for a particle with mass and kinetic energy . Substituting gives:

This shows an inverse relationship between and . Squaring both sides results in:

Graphing vs will yield a hyperbolic curve.
The correct representation is which accurately depicts decreasing as increases, in line with the inverse proportionality.