The wavelengths of electron waves in two orbits are in the ratio $3: 5$. The ratio of the kinetic energy of the electrons will be:

If $a_0$ is denoted as the Bohr radius of the hydrogen atom,then what is the de-Broglie wavelength $(\lambda)$ of the electron present in the $n^{th}$ orbit of the hydrogen atom?

If the de Broglie wavelength of an electron is $728.14 \ nm$,its kinetic energy in $J$ is: (mass of electron $= 9.1 \times 10^{-31} \ kg$; $h = 6.626 \times 10^{-34} \ J \ s$)

The $K.E.$ of an electron is $4.55 \times 10^{-25} \,J$. Calculate its $\lambda$.

The energy of separation of an electron in a $H$ atom in an excited state is $3.4 \ eV$. The de-Broglie wavelength (in $\mathring{A}$) associated with the above electron is,if the radius of the first orbit of the $H$ atom is $0.53 \ \mathring{A}$.

Two particles $A$ and $B$ are in motion. If the wavelength associated with particle $A$ is $5 \times 10^{-8} \ m$,calculate the wavelength (in $\mathring{A}$) associated with particle $B$ if its momentum is half of $A$.