The de-Broglie wavelength of a particle moving with a velocity $2.25 \times 10^8\, m/s$ is equal to the wavelength of a photon. The ratio of the kinetic energy of the particle to the energy of the photon is (velocity of light is $3 \times 10^8\, m/s$). (in $/8$)

After absorbing a slowly moving neutron of mass $m_N$ (momentum $\approx 0$),a nucleus of mass $M$ breaks into two nuclei of masses $m_1$ and $3m_1$ $(4m_1 = M + m_N)$,respectively. If the de Broglie wavelength of the nucleus with mass $m_1$ is $\lambda$,then the de Broglie wavelength of the other nucleus will be:

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

What is the de Broglie wavelength of an electron accelerated through a potential difference of $ 100 \ V $ (in $\text{Å}$)?

The ratio of kinetic energies of a proton and an $\alpha$-particle is $16:1$. What is the ratio of their associated de Broglie wavelengths?

$A$ particle moving with kinetic energy $E$ has de Broglie wavelength $\lambda$. If energy $\Delta E$ is added to its energy,the wavelength becomes $\frac{\lambda}{2}$. The value of $\Delta E$ is: