$A$ 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:

$A$ particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is $1.813 \times 10^{-4}$. Calculate the particle's mass and identify the particle.

If $\lambda_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100 \ V$,the de-Broglie wavelength for an $\alpha$-particle accelerated through the same potential difference is

The accelerating voltage of an electron gun is $50,000 \ V$. What will be the de Broglie wavelength of the electron in $\mathring{A}$?

What is the de Broglie wavelength of
$(a)$ a bullet of mass $0.040 \; kg$ travelling at the speed of $1.0 \; km/s$
$(b)$ a ball of mass $0.060 \; kg$ moving at a speed of $1.0 \; m/s$,and
$(c)$ a dust particle of mass $1.0 \times 10^{-9} \; kg$ drifting with a speed of $2.2 \; m/s$?

$A$ nucleus $A$,with a finite de Broglie wavelength $\lambda_A$,undergoes spontaneous fission into two nuclei $B$ and $C$ of equal mass. $B$ flies in the same direction as that of $A$,while $C$ flies in the opposite direction with a velocity equal to half of that of $B$. The de Broglie wavelengths $\lambda_B$ and $\lambda_C$ of $B$ and $C$ are respectively