If particles move with the same velocity,the de Broglie wavelength is maximum for .........

The de Broglie wavelength associated with an electron accelerated through a potential difference $V$ is $\lambda_e$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_p$. If their corresponding masses are $m_e$ and $m_p$, respectively, then the ratio of their de Broglie wavelengths is . . . . . . .

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?

The kinetic energies of an electron,$\alpha$-particle,and a proton are given as $4K, 2K$,and $K$ respectively. The de-Broglie wavelengths associated with the electron $(\lambda_e)$,$\alpha$-particle $(\lambda_\alpha)$,and the proton $(\lambda_p)$ are related as follows:

Electrons used in an electron microscope are accelerated by a voltage of $25 \ kV$. If the voltage is increased to $100 \ kV$,then the de-Broglie wavelength associated with the electrons would

The de Broglie wavelength associated with an electron accelerated through a potential difference of $\frac{200}{3} \,V$ is nearly (in $Å$)