$A$ proton and an electron are accelerated through the same potential difference. The ratio $\frac{\lambda_e}{\lambda_p}$ will be:

What should be the order of arrangement of de-Broglie wavelength of an electron $(\lambda_{e})$,an $\alpha$-particle $(\lambda_{\alpha})$,and a proton $(\lambda_{p})$,given that all have the same kinetic energy?

$A$ particle $A$ of mass $m$ and charge $q$ is accelerated by a potential difference of $50 \ V$. Another particle $B$ of mass $4m$ and charge $q$ is accelerated by a potential difference of $2500 \ V$. The ratio of de-Broglie wavelength $\frac{\lambda_A}{\lambda_B}$ is close to

$A$ nucleus of mass $M$,moving slowly,absorbs a neutron of mass $m_N$ and then breaks into two nuclei of masses $m_1$ and $5m_1$. If the de Broglie wavelength of the nucleus with mass $m_1$ is $\lambda$,then what will be the de Broglie wavelength of the other nucleus?

The speed of a proton is $c / 20$. What will be the de Broglie wavelength associated with it?

The wavelength of a photon is equal to the wavelength of an electron moving with a kinetic energy of $1.5 \, eV$. What is the energy of the photon?