The ratio of de Broglie wavelengths associated with thermal neutrons at temperatures $127^{\circ} C$ and $352^{\circ} C$ is

If the potential difference used to accelerate electrons is doubled,by what factor does the de Broglie wavelength $(\lambda)$ associated with the electrons change?

An electron of charge $e$ and mass $m$ moving with an initial velocity $v_0 \hat{i}$ is subjected to an electric field $E_0 \hat{j}$. The de-Broglie wavelength of the electron at a time $t$ is (Initial de-Broglie wavelength of the electron $= \lambda_0$)

If $\lambda_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100 \ V$,the de-Broglie wavelength for an $\alpha$-particle accelerated through the same potential difference is

If the mass of a neutron is $1.7 \times 10^{-27} \; kg$,then the de-Broglie wavelength of a neutron with energy $3 \; eV$ is (given $h = 6.6 \times 10^{-34} \; J \cdot s$):

According to the de-Broglie hypothesis,the wavelength associated with a moving electron of mass $m$ is $\lambda_e$. Using the mass-energy relation and Planck's quantum theory,the wavelength associated with a photon is $\lambda_p$. If the energy $(E)$ of the electron and the photon is the same,then the relation between $\lambda_e$ and $\lambda_p$ is: