Correct answer is option A

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

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

where:

• $h$ is Planck's constant,
• $p$ is the momentum of the particle,
• $m$ is the mass of the particle, and
• $K$ is the kinetic energy of the particle.

This equation shows that the de Broglie wavelength of a particle is inversely proportional to the square root of its mass and its kinetic energy. This means that the particle with the smallest mass and kinetic energy will have the largest de Broglie wavelength.

Given that the kinetic energies of the electron, alpha particle, and proton are $4K$, $2K$, and $K$, respectively, and knowing that the mass of the electron is less than the mass of the proton and the mass of the proton is less than the mass of the alpha particle, we can infer that the de Broglie wavelength of the electron is greater than the de Broglie wavelength of the proton, which in turn is greater than the de Broglie wavelength of the alpha particle.

Therefore, the correct answer is option ${\lambda }_{\alpha }<{\lambda }_p<{\lambda }_e$.

Correct answer is C

For same ${\lambda }_1$ momentum should be same,

$(P{)}_P=(P{)}_{\alpha }$

$\Rightarrow \sqrt{2k_P{m}_P}=\sqrt{2k_{\alpha }{m}_{\alpha }}$

$\Rightarrow k_P{m}_P=k_{\alpha }{m}_{\alpha }$

$\Rightarrow \frac{k_P}{k_{\alpha }}=\left(\frac{m_{\alpha }}{m_P}\right)=\frac{4}{1}=4:1$

Correct answer is B

The assertion A and the reason R are both correct statements, and the reason R is the correct Explanation of the assertion A. Explanation The assertion A states that the beam of electrons exhibit wave nature and show interference and diffraction. This statement is correct because electrons exhibit both particle-like and wave-like behavior. When electrons are accelerated to high speeds, they have a wavelength associated with them, and this wavelength can interfere and diffract just like any other wave. The reason R states that the wave nature of electrons was experimentally verified by Davisson Germer. This statement is also correct because Davisson and Germer performed an experiment in 1927 where they observed diffraction patterns in a beam of electrons that were scattered off a nickel crystal. This observation provided strong evidence for the wave nature of electrons. Therefore, both assertion A and reason R are correct statements, and reason R is the correct Explanation:- of assertion A.

Correct answer is option D

$P=\sqrt{2eVm}$

$\lambda =\left(\frac{h}{P_1}\right)$ ..... (i)

$\frac{3\lambda }{2}=\frac{h}{P_2}$ ..... (ii)

Dividing (i) by (ii)

$\Rightarrow \frac{2}{3}=\left(\frac{P_2}{P_1}\right)=\sqrt{\frac{v_2}{v_1}}$

$\Rightarrow \frac{4}{9}=\left(\frac{v_2}{v_1}\right)$

$\frac{v_1}{v_2}=\left(\frac{9}{4}\right)$

Correct answer is option D

Assuming the small object as photon.

Momentum $(p)=\frac{E}{C}$

$=\frac{20\times {10}^{-3}\times 300\times {10}^{-9}}{3\times {10}^8}$

$=2\times {10}^{-17}$ kg m/s