Correct answer is option B

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

where $h$ is Planck's constant, $m$ is the mass of the particle, and $E$ is its kinetic energy.

We are given that the de Broglie wavelength of an electron with kinetic energy $E$ is $\lambda$, and we want to find the de Broglie wavelength of the same electron when its kinetic energy becomes $\frac{E}{4}$.

To do this, we can use the formula for the de Broglie wavelength again, but with the new kinetic energy $\frac{E}{4}$:

${\lambda }^{\prime }=\frac{h}{\sqrt{2m\left(\frac{E}{4}\right)}}=\frac{2h}{\sqrt{2mE}}=2\lambda$

where we have used the fact that $\sqrt{\frac{1}{4}}=\frac{1}{\sqrt{4}}=\frac{1}{2}$ to simplify the expression.

Therefore, the de Broglie wavelength of the electron when its kinetic energy becomes $\frac{E}{4}$ is twice its original value, or $\boxed{{\displaystyle 2\lambda }}$.

Correct answer is option A

The de-Broglie wavelength of a particle is given by:

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

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

The momentum of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by:

$p=\sqrt{2mqV}$

For a proton, $m=1.67\times {10}^{-27}\mathrm{kg}$ and $q=1.6\times {10}^{-19}\mathrm{C}$, and for an $m=6.64\times {10}^{-27}\mathrm{kg}$ and $q=2\times 1.6\times {10}^{-19}\mathrm{C}$.

Therefore, the ratio of their de-Broglie wavelengths is:

$\frac{{\lambda }_p}{{\lambda }_{\alpha }}=\frac{p_{\alpha }}{p_p}=\sqrt{\frac{m_{\alpha }}{m_p}\cdot \frac{q_{\alpha }{V}_{\alpha }}{q_p{V}_p}}=\sqrt{\frac{6.64\times {10}^{-27}}{1.67\times {10}^{-27}}\cdot \frac{2\times 1.6\times {10}^{-19}\times 4}{1.6\times {10}^{-19}\times 2}}=\sqrt{16}=4$

Therefore, the ratio of their de-Broglie wavelengths is 4 : 1.

Correct answer is option A

The de Broglie wavelength (λ) of a particle can be found using the formula:

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

where h is the Planck constant and p is the momentum of the particle. The momentum of a particle can be expressed in terms of its kinetic energy (K) and mass (m) as follows:

$p=\sqrt{2mK}$

Combining these two equations, we get:

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

Now, we are given that the kinetic energy of the proton and electron is the same. Let's denote the masses of the proton and electron as $m_p$ and $m_e$, respectively. We are given the relationship between the two masses:

$m_p=1849\times m_e$

Let's find the ratio of the de Broglie wavelengths of the proton (${\lambda }_p$) and the electron (${\lambda }_e$):

$\frac{{\lambda }_p}{{\lambda }_e}=\frac{\frac{h}{\sqrt{2m_{p}K}}}{\frac{h}{\sqrt{2m_{e}K}}}$

Simplifying the expression, we get:

$\frac{{\lambda }_p}{{\lambda }_e}=\frac{\sqrt{2m_{e}K}}{\sqrt{2m_{p}K}}$ The 2K terms cancel out:

$\frac{{\lambda }_p}{{\lambda }_e}=\frac{\sqrt{m_e}}{\sqrt{m_p}}$

Substitute the given relationship between the masses:

$\frac{{\lambda }_p}{{\lambda }_e}=\frac{\sqrt{m_e}}{\sqrt{1849\times m_e}}$

Further simplification:

$\frac{{\lambda }_p}{{\lambda }_e}=\frac{1}{\sqrt{1849}}$

Since 1849 is equal to ${43}^2$:

$\frac{{\lambda }_p}{{\lambda }_e}=\frac{1}{43}$

Thus, the ratio of the de Broglie wavelengths of the proton and electron having the same kinetic energy is 1:43.

Correct answer is option C

The de Broglie wavelength of a particle is given by:

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

where h is Planck's constant and p is the momentum of the particle. The momentum of a gas molecule can be related to its kinetic energy (which is related to the temperature of the gas) by:

$p=\sqrt{2mK}$

where m is the mass of the molecule and K is the kinetic energy of the molecule.

At a given temperature T, the average kinetic energy of a molecule in a gas is given by:

$K=\frac{3}{2}kT$

where k is Boltzmann's constant.

Therefore, the de Broglie wavelength of a molecule in a gas is given by:

$\lambda =\frac{h}{\sqrt{2m(3/2)kT}}=\frac{h}{\sqrt{3mkT}}$

If the temperature of the gas is increased from T = 300 K to T = 600 K, the new de Broglie wavelength becomes:

${\lambda }^{\prime }=\frac{h}{\sqrt{3mk(2T)}}=\frac{h}{\sqrt{2}\sqrt{3mkT}}=\frac{1}{\sqrt{2}}\lambda$

So, the de Broglie wavelength of the gas molecule decreases by a factor of $\sqrt{2}$ when the temperature of the gas is doubled.

Therefore, the correct answer is option ${\lambda }^{\prime }=\frac{1}{\sqrt{2}}{\lambda }_1$

Correct answer is option A

The de Broglie wavelength of a particle is given by the formula:

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

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

If the de Broglie wavelengths of the proton and electron are the same, then:

$\frac{h}{p_p}=\frac{h}{p_e}$

where $p_p$ and $p_e$ are the momenta of the proton and electron, respectively.

Solving this equation for the ratio of their momenta gives:

$\frac{p_p}{p_e}=1$

So, the ratio of their momenta is 1:1