The kinetic energy of an electron with a de-Broglie wavelength of $0.3 \, nm$ is ............. $eV$.

An $\alpha$-particle and a deuteron are moving with velocities $v$ and $2v$ respectively. What will be the ratio of their de Broglie wavelengths?

Assuming the nitrogen molecule is moving with $r.m.s.$ velocity at $400 \ K$, the de$-$Broglie wavelength of the nitrogen molecule is close to $...... \ \mathring{A}$. (Given: nitrogen molecule mass: $4.64 \times 10^{-26} \ kg$, Boltzmann constant: $1.38 \times 10^{-23} \ J/K$, Planck constant: $6.63 \times 10^{-34} \ J \cdot s$)

The de Broglie wavelength $(\lambda)$ depends on mass '$m$' and kinetic energy '$E$' according to which formula?

The de Broglie wavelength of a proton accelerated by a potential difference of $100 \ V$ is $\lambda_0$. If an alpha particle is accelerated by the same potential difference,its de Broglie wavelength will be:

The de Broglie wavelength of an oxygen molecule at $27^{\circ} C$ is $x \times 10^{-12} \ m$. The value of $x$ is (take Planck's constant $= 6.63 \times 10^{-34} \ J \cdot s$,Boltzmann constant $= 1.38 \times 10^{-23} \ J/K$,mass of oxygen molecule $= 5.31 \times 10^{-26} \ kg$).