The equation $\lambda = \frac{1.227}{x} \text{ nm}$ can be used to find the de-Broglie wavelength of an electron. In this equation,$x$ stands for:

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with electrons change?

The velocity of a particle $A$ is $3$ times the velocity of a proton. If the ratio of the de Broglie wavelengths of the particle $A$ and the proton is $3:2$,the mass of the particle $A$ is (where $m_{p}$ is the mass of the proton).

The de-Broglie wavelength of an electron accelerated by a potential of $50\,V$ is approximately ............. $\mathring{A}$. $(|e| = 1.6 \times 10^{-19}\,C, m_e = 9.1 \times 10^{-31}\,kg, h = 6.6 \times 10^{-34}\,Js)$

$A$ nucleus $A$,with a finite de Broglie wavelength $\lambda_A$,undergoes spontaneous fission into two nuclei $B$ and $C$ of equal mass. $B$ flies in the same direction as that of $A$,while $C$ flies in the opposite direction with a velocity equal to half of that of $B$. The de Broglie wavelengths $\lambda_B$ and $\lambda_C$ of $B$ and $C$ are respectively

De Broglie waves are associated with moving particles. These particles can be .......