If the de-Broglie wavelength at a temperature of $27^oC$ is $\lambda$,what will be the de-Broglie wavelength at $927^oC$?

An electron with an initial kinetic energy of $100 \, eV$ is accelerated through a potential difference of $50 \, V$. The final de-Broglie wavelength of the electron becomes .................... $\mathring{A}$

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

The de-Broglie wavelength of a particle accelerated with $150 \ V$ potential is $10^{-10} \ m$. If it is accelerated by $600 \ V$ potential difference,its wavelength will be: ............... $\mathring{A}$

$A$ photon and an electron (mass $m$) have the same energy $E$. The ratio $\left(\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}}\right)$ of their de Broglie wavelengths is: ($c$ is the speed of light)

An electron of mass $m$ and charge $e$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of the electron at time $t$ is (Ignore relativistic effect) ($h=$ Planck's constant).