An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength $\lambda$. When the potential is changed to $V_2$,its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{V_1}{V_2}\right)$ is

When energy is added to an electron,its de Broglie wavelength decreases from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$. The added energy is:

The interatomic spacing in a crystal is $1.227 \ \mathring A$. What is the maximum order of diffraction for electrons accelerated by $10 \ kV$?

Crystal diffraction experiments can be performed using $X-$rays,or electrons accelerated through an appropriate voltage. Which probe has greater energy? (For quantitative comparison,take the wavelength of the probe equal to $1\,\mathring{A}$,which is of the order of inter-atomic spacing in the lattice) $(m_{e}=9.11 \times 10^{-31}\,kg)$

The kinetic energy of an electron with a de-Broglie wavelength of $0.3 \, nm$ is ............. $eV$.

$A$ proton and an $\alpha$-particle are accelerated from rest by $2\,V$ and $4\,V$ potentials, respectively. The ratio of their de-Broglie wavelength is: (in $:1$)