The de-Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of $3 \ km/s$ will be ............ $\mathring{A}$.

An electron microscope uses electrons of $40 \ keV$. The de Broglie wavelength associated with these electrons is approximately:

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

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of $n$ is

$A$ 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

Write the equation for the de-Broglie wavelength of a particle having momentum $(p)$.