If the total energy of an electron in a hydrogen atom in an excited state is $-3.4 \ eV$,then the de-Broglie wavelength of the electron 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 $\left(m_e = 9 \times 10^{-31} \ kg\right)$.

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

What accelerating potential must be imparted to a proton beam to give it an effective wavelength of $\lambda = 0.05 \ \mathring{A}$? (Given: $m_p = 1.672 \times 10^{-27} \ kg$,$h = 6.626 \times 10^{-34} \ J \cdot s$,$e = 1.602 \times 10^{-19} \ C$)

Which is the correct relationship between wavelength and momentum of particles?

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