If the de Broglie wavelength of a dust particle of mass $1.0 \times 10^{-9} \,kg$ is $3 \times 10^{-25} \,m$, then the speed of the particle is . . . . . . . $\left(h=6.625 \times 10^{-34} \,J \,s\right)$

Electrons are accelerated through a potential difference $V$ and protons are accelerated through a potential difference $4V$. The de-Broglie wavelengths are $\lambda_e$ and $\lambda_p$ for electrons and protons respectively. The ratio of $\frac{\lambda_e}{\lambda_p}$ is given by: (given $m_e$ is the mass of the electron and $m_p$ is the mass of the proton).

$A$ photosensitive metallic surface emits electrons when $X$-rays of wavelength $\lambda$ fall on it. The de-Broglie wavelength of the emitted electrons is (Neglect the work function of the surface,$m$ is mass of the electron,$h$ is Planck's constant,$c$ is the velocity of light).

If a proton and an $\alpha$-particle have the same kinetic energy,what is the ratio of their de-Broglie wavelengths?

The accelerating voltage of an electron gun is $50,000 \ V$. What will be the de Broglie wavelength of the electron in $\mathring{A}$?

$A$ charged particle $q_0$ of mass $m_0$ is projected along the $y-$axis at $t = 0$ from the origin with a velocity $v_0$. If a uniform electric field $E_0$ also exists along the $x-$axis,then the time at which the de Broglie wavelength of the particle becomes half of its initial value is: