With what potential an electron should be accelerated so that its de Broglie wavelength becomes equal to the wavelength of the first line of the Lyman series for the $He^+$ ion?

$A$ deuteron is accelerated through $500 \ V$. For the same de Broglie wavelength,a singly ionized helium ion is accelerated through a potential difference of $V \ V$. Find the value of $V$.

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?

The relation between the wavelength of electromagnetic radiation $(\lambda)$ and the de Broglie wavelength of its quantum (photon) $(\lambda')$ is . . . . . . .

An electron with mass $m$ and an initial velocity $(t=0)$ $\vec{v} = v_0 \hat{i}$ $(v_0 > 0)$ enters a magnetic field $\vec{B} = B_0 \hat{j}$. If the initial de Broglie wavelength at $t=0$ is $\lambda_0$,then its value after time $t$ would be:

If a proton and an alpha particle are accelerated through the same potential difference, then the ratio of their de Broglie wavelengths is