$A$ photon and an electron have equal energy $E$. The ratio $\lambda_{\text{photon}} / \lambda_{\text{electron}}$ is proportional to:

$A$ stationary mass $M$ splits into two fragments of masses $m_1$ and $m_2$. What is the ratio of their de Broglie wavelengths,${\lambda _1}/{\lambda _2}$?

An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is

If an electron has an energy such that its de-Broglie wavelength is $5500 \ \text{Å}$,then the energy value of that electron is $(h = 6.6 \times 10^{-34} \ \text{Js}, m_e = 9.1 \times 10^{-31} \ \text{kg})$.

$A$ 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 $:1$)

The de Broglie wavelength for a deuteron can be given by: