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

$A$ proton and an $\alpha$-particle are accelerated through the same potential difference $V$. The ratio of their de-Broglie wavelengths is

The ratio of wavelengths of a proton and a deuteron accelerated by potentials $V_{p}$ and $V_{d}$ is $1 : \sqrt{2}$. Then,the ratio of $V_{p}$ to $V_{d}$ will be:

$A$ nucleus of mass $M$ at rest splits into two parts having masses $\frac{M'}{3}$ and $\frac{2M'}{3}$ (where $M' < M$). The ratio of the de Broglie wavelengths of the two parts will be:

An electron of mass $m_{e}$ and a proton of mass $m_{p} = 1836 m_{e}$ are moving with the same speed. The ratio of their de Broglie wavelength $\frac{\lambda_{\text{electron}}}{\lambda_{\text{proton}}}$ will be ....... .

The de-Broglie wavelength of a proton (charge $= 1.6 \times 10^{-19} \ C$,mass $= 1.67 \times 10^{-27} \ kg$) accelerated through a potential difference of $1 \ kV$ is: