Find the de Broglie wavelength associated with a Helium $(He)$ atom in Helium gas at room temperature $(27^{\circ}C)$ and $1$ atmospheric pressure.

If the momentum of an electron is changed by $\Delta p,$ then the de-Broglie wavelength associated with it changes by $0.50\%.$ The initial momentum of the electron will be:

If particles are moving with the same velocity,then the maximum de-Broglie wavelength will be for

Electrons are accelerated through a potential difference $V$ and protons are accelerated through a potential difference $4V$. The de-Broglie wavelengths are $\lambda_e$ and $\lambda_p$ for electrons and protons respectively. The ratio of $\frac{\lambda_e}{\lambda_p}$ is given by: (given $m_e$ is the mass of the electron and $m_p$ is the mass of the proton).

An electron of mass $m$ and a photon have the same energy $E$. The ratio of the de-Broglie wavelengths associated with them is:

If the kinetic energy of an electron is increased four times,the wavelength of the de-Broglie wave associated with it would become