The de-Broglie wavelength of a body of mass $1 \ kg$ moving with a velocity of $2000 \ m/s$ is:

The graph which shows the variation of $\left(\frac{1}{\lambda^2}\right)$ and its kinetic energy,$E$ is (where $\lambda$ is de Broglie wavelength of a free particle):

$A$ proton is accelerated through $225 \,V$. Its de Broglie wavelength is ........ $nm$.

The kinetic energies of an electron,$\alpha$-particle,and a proton are given as $4K, 2K$,and $K$ respectively. The de-Broglie wavelengths associated with the electron $(\lambda_e)$,$\alpha$-particle $(\lambda_\alpha)$,and the proton $(\lambda_p)$ are related as follows:

The de Broglie wavelength of an electron of kinetic energy $9 \ eV$ is (take $h=4 \times 10^{-15} \ eV \cdot s$,$c=3 \times 10^{10} \ cm/s$ and the mass $m_e$ of electron as $m_e c^2=0.5 \ MeV$)

$A$ photon and an electron (mass $m$) have the same energy $E$. The ratio $\left(\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}}\right)$ of their de Broglie wavelengths is: ($c$ is the speed of light)