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

$A$ particle has a de Broglie wavelength $\lambda$. If its velocity and energy are both doubled,what will be the new de Broglie wavelength?

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

If the kinetic energy of a moving particle is $E$,then the De Broglie wavelength is

Calculate the de Broglie wavelength of an electron with a kinetic energy of $200 \ eV$. [Mass of electron $= 1 \times 10^{-30} \ kg$,charge on electron $= 1.6 \times 10^{-19} \ C$,Planck's constant $(h) = 6.6 \times 10^{-34} \ Js$].

$A$ nucleus of mass $M$,moving slowly,absorbs a neutron of mass $m_N$ and then breaks into two nuclei of masses $m_1$ and $5m_1$. If the de Broglie wavelength of the nucleus with mass $m_1$ is $\lambda$,then what will be the de Broglie wavelength of the other nucleus?