The graph which shows the variation of the de Broglie wavelength $(\lambda)$ of a particle and its associated momentum $(p)$ is

If the kinetic energy of a particle is increased by $16$ times, the percentage change in the de Broglie wavelength of the particle is......... $\%$

$A$ free particle with initial kinetic energy $E$ and de-Broglie wavelength $\lambda$ enters a region in which it has potential energy $V$. What is the particle's new de-Broglie wavelength?

$A$ proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

If we consider an electron and a photon of the same de-Broglie wavelength,then they will have the same:

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is ($c$ = velocity of light in air).