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 ratio of de Broglie wavelengths of a proton and an $\alpha$-particle having the same energy is ............

The graph which shows the variation of the de Broglie wavelength $(\lambda)$ of a particle and its associated momentum $(p)$ is

$A$ proton and an electron are accelerated by the same potential difference. If their de-Broglie wavelengths are $\lambda_p$ and $\lambda_e$ respectively,then:

The ratio of the de-Broglie wavelengths of a proton and an electron having the same kinetic energy is: (Assume $m_p = 1849 \times m_e$)

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: