If the velocity of a hydrogen molecule is $5 \times 10^4 \, cm \, sec^{-1}$,then its de-Broglie wavelength is ............. $\mathring{A}$.

In the ground state of a hydrogen atom,an electron absorbs $1.5$ times the minimum energy $\left(2.18 \times 10^{-18} \ J\right)$ required to escape from the atom. The wavelength of the emitted electron (in $m$) is $\left(m_e = 9 \times 10^{-31} \ kg\right)$.

If the kinetic energy of an electron is $18.2 \times 10^{-25} \ J$,its de Broglie wavelength in $nm$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

State the de-Broglie principle and explain it.

The de Broglie wavelength of a particle $X$ moving with a certain velocity is $1 \ \mathring{A}$. If the mass of $Y$ is $25\%$ of $X$ and its velocity is $75\%$ of $X$,then the de Broglie wavelength of $Y$ will be ........... $\mathring{A}$.

What is the characteristic of an electron microscope? On which principle is it based?