An $\alpha$ particle and a carbon $12$ atom have the same kinetic energy $K$. The ratio of their de-Broglie wavelengths $(\lambda_{\alpha} : \lambda_{C12})$ is

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

$A$ 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 $\mathring{A}$.

$A$ proton moving with one-tenth of the velocity of light has a certain de Broglie wavelength of $\lambda$. An alpha particle having a certain kinetic energy has the same de Broglie wavelength $\lambda$. The ratio of the kinetic energy of the proton to that of the alpha particle is:

When the kinetic energy of an electron is increased,the wavelength of the associated wave will:

Consider an electron $(m = 9.1 \times 10^{-31} \ kg)$ confined by electrical forces to move between two rigid walls separated by $1.0 \times 10^{-9} \ m$,which is about five atomic diameters. The quantized energy value for the lowest stationary state is