If the kinetic energy of a particle is increased to $16$ times, the percentage change in the de Broglie wavelength of a particle is (in $\%$)

If the kinetic energy of a free electron doubles,its de-Broglie wavelength changes by the factor

$A$ beam of electrons of energy $E$ scatters from a target having atomic spacing of $1 \, Å$. The first maximum intensity occurs at $\theta = 60^{\circ}$. Then $E$ (in $eV$) is: (Planck constant $h = 6.64 \times 10^{-34} \, Js$, $1 \, eV = 1.6 \times 10^{-19} \, J$, electron mass $m = 9.1 \times 10^{-31} \, kg$)

The de Broglie wavelength and kinetic energy of a particle are $2000 \ \mathring{A}$ and $1 \ \text{eV}$ respectively. If its kinetic energy becomes $1 \ \text{MeV}$,then its de Broglie wavelength becomes $...... \ \mathring{A}$.

An $\alpha$-particle and a proton are accelerated from rest by the same potential. The ratio of their de-Broglie wavelengths is . . . . . .

If we consider an electron and a photon of the same de-Broglie wavelength,then they will have the same: