$A$ proton when accelerated through a potential difference of $V$, has a de-Broglie wavelength $\lambda$ associated with it. If an $\alpha$-particle is to have the same de-Broglie wavelength $\lambda$, it must be accelerated through a potential difference of

An $\alpha$-particle and a deuteron are moving with velocities $v$ and $2v$ respectively. What will be the ratio of their de Broglie wavelengths?

$A$ ball of mass $0.12 \ kg$ is moving with a speed $20 \ m \ s^{-1}$. Then its de Broglie wavelength is . . . . . . . ( $h = 6.63 \times 10^{-34} \ J \ s$ )

The de Broglie wavelength of an electron of kinetic energy $9 \ eV$ is (take $h=4 \times 10^{-15} \ eV \cdot s$,$c=3 \times 10^{10} \ cm/s$ and the mass $m_e$ of electron as $m_e c^2=0.5 \ MeV$)

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:

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).