Correct option is (B)

To determine the relationship between the de-Broglie wavelengths of a proton, an electron, and an alpha particle with the same energies, we need to use the de-Broglie wavelength formula:

$\lambda =\frac{h}{p}$

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

For particles with the same kinetic energy $E$, we have:

$E=\frac{p^2}{2m}$

Solving for $p$:

$p=\sqrt{2mE}$

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

$\lambda =\frac{h}{\sqrt{2mE}}$

Since all three particles have the same energy $E$, the de-Broglie wavelength is inversely proportional to the square root of the mass $m$:

$\lambda \propto \frac{1}{\sqrt{m}}$

The masses of the particles are as follows:

• Mass of proton $m_p$ is approximately $1.67\times {10}^{-27}$ kg
• Mass of electron $m_e$ is approximately $9.11\times {10}^{-31}$ kg
• Mass of alpha particle $m_{\alpha }$ is approximately $4\times 1.67\times {10}^{-27}$ kg (since it has 2 protons and 2 neutrons)

Comparatively:

• Mass of alpha particle $m_{\alpha }$ is the largest.
• Mass of proton $m_p$ is intermediate.
• Mass of electron $m_e$ is the smallest.

Therefore, the de-Broglie wavelength will be:

• Largest for the electron (${\lambda }_{\mathrm{e}}$)
• Intermediate for the proton (${\lambda }_{\mathrm{p}}$)
• Smallest for the alpha particle (${\lambda }_{\alpha }$)

Hence, the correct order is:

Option B:

${\lambda }_{\alpha }<{\lambda }_{\mathrm{p}}<{\lambda }_{\mathrm{e}}$

Correct option is (C)

To determine the correct relation between the kinetic energies of a proton (${\mathrm{K}}_{\mathrm{p}}$) and an electron (${\mathrm{K}}_{\mathrm{e}}$) when they have the same de Broglie wavelength, we need to use the de Broglie wavelength formula:

$\lambda =\frac{h}{p}$

where $\lambda$ is the de Broglie wavelength, $h$ is Planck's constant, and $p$ is the momentum of the particle.

The momentum $p$ of a particle is given by:

$p=\sqrt{2mK}$

where $m$ is the mass of the particle and $K$ is its kinetic energy. For a proton and an electron with the same de Broglie wavelength:

${\lambda }_{\mathrm{p}}={\lambda }_{\mathrm{e}}$

This implies the momenta should be the same:

$p_{\mathrm{p}}=p_{\mathrm{e}}$

Thus, we can write:

$\sqrt{2m_{\mathrm{p}}{K}_{\mathrm{p}}}=\sqrt{2m_{\mathrm{e}}{K}_{\mathrm{e}}}$

Squaring both sides to eliminate the square roots:

$2m_{\mathrm{p}}{K}_{\mathrm{p}}=2m_{\mathrm{e}}{K}_{\mathrm{e}}$

$m_{\mathrm{p}}{K}_{\mathrm{p}}=m_{\mathrm{e}}{K}_{\mathrm{e}}$

Rearranging to solve for $K_{\mathrm{p}}$ in terms of $K_{\mathrm{e}}$:

$K_{\mathrm{p}}=\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}}{K}_{\mathrm{e}}$

Since the mass of a proton $m_{\mathrm{p}}$ is much greater than the mass of an electron $m_{\mathrm{e}}$:

$m_{\mathrm{p}}\gg m_{\mathrm{e}}$

This means:

$\frac{m_{\mathrm{e}}}{m_{\mathrm{p}}}\ll 1$

Therefore, it implies:

$K_{\mathrm{p}}<K_{\mathrm{e}}$

So, the correct option is:

Option C: ${\mathrm{K}}_{\mathrm{p}}<{\mathrm{K}}_{\mathrm{e}}$

Correct option is (C)

To solve for the ratio of the kinetic energies of a proton and an electron with the same de-Broglie wavelength, let us first recall the relationship between kinetic energy, momentum, and the de-Broglie wavelength.

The de-Broglie wavelength $\lambda$ is given by:

$\lambda =\frac{h}{p}$

where $h$ is Planck’s constant and $p$ is the momentum.

Since the proton and the electron have the same de-Broglie wavelength, their momenta must be equal:

${\lambda }_{\text{electron}}={\lambda }_{\text{proton}}⟹\frac{h}{p_{\text{e}}}=\frac{h}{p_{\text{p}}}⟹p_{\text{e}}=p_{\text{p}}$

Next, the kinetic energy $K$ of a particle is related to its momentum $p$ and mass $m$ by the following formula:

$K=\frac{p^2}{2m}$

Given that the momentum $p$ is the same for both the proton and the electron, we can write the kinetic energies as:

$K_{\text{e}}=\frac{p^2}{2m_{\text{e}}}$

$K_{\text{p}}=\frac{p^2}{2m_{\text{p}}}$

The ratio of the kinetic energies is therefore:

$\frac{K_{\text{e}}}{K_{\text{p}}}=\frac{\frac{p^2}{2m_{\text{e}}}}{\frac{p^2}{2m_{\text{p}}}}=\frac{m_{\text{p}}}{m_{\text{e}}}$

Given that the mass of the proton $m_{\text{p}}$ is 1836 times the mass of the electron $m_{\text{e}}$, we have:

$\frac{m_{\text{p}}}{m_{\text{e}}}=1836$

Thus, the ratio of the kinetic energies is:

$\frac{K_{\text{e}}}{K_{\text{p}}}=1836$

Therefore, the correct answer is:

Option C

$1:1836$

Correct option is (A)

To find the ratio of velocities of two particles based on their de Broglie wavelengths, we can use the de Broglie wavelength formula, which relates the momentum of a particle to its wavelength. The de Broglie's wavelength formula is given by:

$\lambda =\frac{h}{p}$

where:

$\lambda$ is the de Broglie wavelength,

$h$ is the Planck constant, and

$p$ is the momentum of the particle.

The momentum $p$ of a particle can also be expressed as the product of its mass $m$ and velocity $v$:

$p=mv$

So, we can rewrite the de Broglie wavelength equation in terms of mass and velocity:

$\lambda =\frac{h}{mv}$

For the proton (let's use subscript $p$ for proton), the wavelength is $\lambda$:

${\lambda }_p=\frac{h}{m_p{v}_p}$

For the $\alpha$ particle (let's use subscript $\alpha$ for alpha), the wavelength is $2\lambda$:

$2\lambda =\frac{h}{m_{\alpha }{v}_{\alpha }}$

We are interested in finding the ratio of the velocities $\frac{v_p}{v_{\alpha }}$. Using the given data about wavelengths:

${\lambda }_p=\lambda$

${\lambda }_{\alpha }=2\lambda$

Using the de Broglie equation for both particles:

$\frac{h}{m_p{v}_p}=\lambda$

$\frac{h}{m_{\alpha }{v}_{\alpha }}=2\lambda$

Dividing the second equation by the first equation gives us:

$\frac{\frac{h}{m_{\alpha }{v}_{\alpha }}}{\frac{h}{m_p{v}_p}}=\frac{2\lambda }{\lambda }$

$\frac{m_p{v}_p}{m_{\alpha }{v}_{\alpha }}=\frac{2}{1}$

$\frac{v_p}{v_{\alpha }}=\frac{2m_{\alpha }}{m_p}$

Since we know an $\alpha$ particle consists of 2 protons and 2 neutrons (essentially four nucleons), the mass of an $\alpha$ particle is roughly four times the mass of a proton ($m_{\alpha }\approx 4m_p$).

Substituting $m_{\alpha }$ with $4m_p$ in the equation:

$\frac{v_p}{v_{\alpha }}=\frac{2(4m_p)}{m_p}$

$\frac{v_p}{v_{\alpha }}=2\cdot 4$

$\frac{v_p}{v_{\alpha }}=8$

Therefore, the ratio of the velocities of proton to $\alpha$ particle is $8:1$, which corresponds to Option A.

Correct option is (C)

We know that the de-Broglie wavelength $\lambda$ of a particle is given by:

$\lambda =\frac{h}{p}$

where:

• $h$ is Planck's constant,
• $p$ is the momentum of the particle.

For a photon (which has zero rest mass), its energy $E$ and momentum $p$ are related by the equation:

$E=cp$

and its de-Broglie wavelength $\lambda$ is given by:

$\lambda =\frac{h}{E}\times \frac{1}{c}$

Now, since the photon and electron are said to have the same de-Broglie wavelength:

${\lambda }_{electron}={\lambda }_{photon}$

$\frac{h}{p_{electron}}=\frac{h}{E_{photon}}\times \frac{1}{c}$

For the electron, its momentum $p_{electron}$ is given by:

$p_{electron}=m_e{v}_{electron}$

where:

• $m_e$ is the electron's rest mass,
• $v_{electron}$ is the electron's velocity.

The kinetic energy $K.E.$ of the electron is:

$K.E{.}_{electron}=\frac{1}{2}{m}_e{v}_{electron}^2$

The question states that $v_{electron}$ is $25\mathrm{\%}$ ($0.25c$) of the speed of light $c$. So we write:

$v_{electron}=0.25c$

Plugging this into the kinetic energy formula, we get:

$K.E{.}_{electron}=\frac{1}{2}{m}_e(0.25c{)}^2=\frac{1}{2}{m}_e\times \frac{1}{16}{c}^2$

$K.E{.}_{electron}=\frac{1}{32}{m}_e{c}^2$

For a photon, $p_{photon}=\frac{E_{photon}}{c}$ and hence its kinetic energy (which, for a photon, is simply its energy) is:

$K.E{.}_{photon}=E_{photon}=c{p}_{photon}$

Now, comparing the kinetic energies:

$\frac{K.E{.}_{electron}}{K.E{.}_{photon}}=\frac{\frac{1}{32}{m}_e{c}^2}{c{p}_{photon}}$

Since $p_{electron}=p_{photon}$ (from the de-Broglie relation), we can replace $p_{photon}$ with $p_{electron}$ which is $m_e{v}_{electron}$:

$\frac{K.E{.}_{electron}}{K.E{.}_{photon}}=\frac{\frac{1}{32}{m}_e{c}^2}{m_e{v}_{electron}c}$

$\frac{K.E{.}_{electron}}{K.E{.}_{photon}}=\frac{\frac{1}{32}c}{0.25c}$

$\frac{K.E{.}_{electron}}{K.E{.}_{photon}}=\frac{1}{32}\times \frac{1}{0.25}$

$\frac{K.E{.}_{electron}}{K.E{.}_{photon}}=\frac{1}{32}\times 4$

$\frac{K.E{.}_{electron}}{K.E{.}_{photon}}=\frac{1}{8}$

So the correct answer is Option C $\frac{1}{8}$.