If the kinetic energy of an electron of mass $9.0 \times 10^{-31} \ kg$ is $8.0 \times 10^{-25} \ J$,the wavelength of this electron in $nm$ is

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

Calculate the wavelength (in nanometer) associated with a proton moving at $1.0 \times 10^3 \ m \ s^{-1}$. (Mass of proton $= 1.67 \times 10^{-27} \ kg$ and $h = 6.63 \times 10^{-34} \ J \ s$)

If the kinetic energy of a particle is reduced to half,the de-Broglie wavelength becomes

If the de Broglie wavelength of the electron in the $n^{th}$ Bohr orbit in a hydrogenic atom is equal to $1.5 \, \pi \, a_0$ ($a_0$ is Bohr radius),then the value of $n/Z$ is:

If $v_0$ is the threshold frequency of a metal $X$,the correct relation between de Broglie wavelength $(\lambda)$ associated with the photoelectron and the frequency $(v)$ of the incident radiation is: