The de-Broglie wavelength of a particle having kinetic energy $E$ is $\lambda$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to $75 \%$ of the initial value?

The graph between the energy log $E$ of an electron and its de$-$Broglie wavelength log $\lambda$ will be

The de Broglie wavelength of an oxygen molecule at $27^{\circ} C$ is $x \times 10^{-12} \ m$. The value of $x$ is (take Planck's constant $= 6.63 \times 10^{-34} \ J \cdot s$, Boltzmann constant $= 1.38 \times 10^{-23} \ J/K$, mass of oxygen molecule $= 5.31 \times 10^{-26} \ kg$).

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ is:

The de-Broglie wavelength of an electron having kinetic energy $100 eV$ is, $[$ Use $h=4.14 \times 10^{-15} eVs$, mass of electron $= \frac{0.5 \times 10^6}{c^2} eV/c^2$, $1 pm = 10^{-12} m$ $]$ (in $pm$)

$A$ proton,a neutron,an electron,and an $\alpha$-particle have the same energy. If $\lambda_{p}, \lambda_{n}, \lambda_{e},$ and $\lambda_{\alpha}$ are the de Broglie wavelengths of the proton,neutron,electron,and $\alpha$-particle respectively,then choose the correct relation from the following: