What is the additional energy that should be supplied to a moving electron to reduce its de Broglie wavelength from $1 \,nm$ to $0.5 \,nm$?

The ratio of the de-Broglie wavelengths of hydrogen and helium atoms at $27^{\circ}C$ is .....

The graph shows the variation of de Broglie wavelength $(\lambda)$ versus $\frac{1}{\sqrt{V}}$,where '$V$' is the accelerating potential for four particles $A, B, C, D$ carrying the same charge but having masses $m_1, m_2, m_3, m_4$. Which one represents a particle of the largest mass?

$A$ proton and a photon both have the same energy of $E = 100 \,keV$. If the de Broglie wavelength of the proton is $\lambda_1$ and that of the photon is $\lambda_2$,then $\lambda_1/\lambda_2$ is proportional to:

An electron of mass $m$ and magnitude of charge $|e|$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is:

$A$ photon and an electron have equal energy $E$. The ratio $\lambda_{\text{photon}} / \lambda_{\text{electron}}$ is proportional to: