$A$ proton moves in a circular path of radius $6.6 \times 10^{-3} \ m$ perpendicular to a magnetic field of $0.625 \ T$. Find the de Broglie wavelength associated with the proton in $\mathring{A}$.

$A$ nucleus of mass $M$,moving slowly,absorbs a neutron of mass $m_N$ and then breaks into two nuclei of masses $m_1$ and $5m_1$. If the de Broglie wavelength of the nucleus with mass $m_1$ is $\lambda$,then what will be the de Broglie wavelength of the other nucleus?

The de-Broglie wavelength of an electron moving with a velocity $v = c / 2$ ($c$ is the velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of the electron and the photon is:

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

The de Broglie wavelength of an electron moving with kinetic energy of $144 \;eV$ is nearly

The de Broglie wavelength associated with an electron accelerated through a potential difference of $\frac{200}{3} \,V$ is nearly (in $Å$)