Calculate the de-Broglie wavelength of an electron residing in the $2$nd Bohr orbit of a hydrogen atom. (Bohr radius,$a_0 = 0.529 \ \mathring{A}$)

The ratio of de Broglie wavelengths of two particles,having mass ratio $1 : 3$ and kinetic energy ratio $2 : 1$ is

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 $\left(m_e = 9 \times 10^{-31} \ kg\right)$.

The atomic masses of $He$ and $Ne$ are $4$ and $20 \ amu$,respectively. The value of the de Broglie wavelength of $He$ gas at $-73^{\circ} C$ is $M$ times that of the de Broglie wavelength of $Ne$ at $727^{\circ} C$. $M$ is

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

What is the de Broglie wavelength of a tennis ball weighing $66 \ g$ moving with a velocity of $10 \ m/s$?