The de-Broglie wavelength of a body of mass $m$ and kinetic energy $E$ is given by:

$A$ particle of mass $2 \times 10^{-27} \,kg$ has a de-Broglie wavelength of $3.3 \times 10^{-10} \,m$. The kinetic energy of this particle is (Planck's constant $h = 6.6 \times 10^{-34} \,J \cdot s$).

The de Broglie wavelength of an electron accelerated to a potential of $ 400 \,V $ is approximately (in $\,nm$)

Kinetic energy of a proton is equal to energy $E$ of a photon. Let $\lambda_1$ be the de-Broglie wavelength of the proton and $\lambda_2$ be the wavelength of the photon. If $\frac{\lambda_1}{\lambda_2} \propto E^{n}$,then the value of $n$ is

$A$ 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

The de Broglie wavelength for an electron accelerated through a potential difference of $V_1$ volt is $\lambda_1$. When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by $50\%$. If $(V_1/V_2) = (9/\alpha)$, then the value of $\alpha$ is . . . . . . .