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

If an electron and a photon propagate in the form of waves having the same wavelength,it implies that they have the same

Choose the only correct statement out of the following:

If the kinetic energy of a moving particle is $E$,then the De Broglie wavelength is

$A$ photon and an electron have equal energy $E$. The ratio of $\lambda(\text{electron})$ to $\lambda(\text{photon})$ is proportional to

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: