The ratio of the de-Broglie wavelength of molecules of hydrogen $(H_2)$ and helium $(He)$ which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is:

The ratio of momenta of an electron and an $\alpha$-particle which are accelerated from rest by a potential difference of $100 \ V$ is

The de Broglie wavelengths for an electron and a photon are $\lambda_{e}$ and $\lambda_{p}$ respectively. For the same kinetic energy $K$ of the electron and the photon,which of the following presents the correct relation between the de Broglie wavelengths of the two?

The kinetic energy of an electron is $5 \ eV$. Calculate the de-Broglie wavelength associated with it in $\mathring{A}$. $(h = 6.6 \times 10^{-34} \ J \cdot s, m_e = 9.1 \times 10^{-31} \ kg)$

An electron of mass $m$ and magnitude of charge $|e|$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is:

The de-Broglie wavelength associated with an electron,accelerated through a potential difference of $121 \ V$ is about:
[Take Planck's constant $h = 6.6 \times 10^{-34} \ J \cdot s$,mass of electron $m = 9 \times 10^{-31} \ kg$,charge of electron $e = 1.6 \times 10^{-19} \ C$] (in $nm$)