The kinetic energy of a free electron increases to $3$ times the previous kinetic energy $(K.E.)$. The ratio of the new de-Broglie wavelength to the previous de-Broglie wavelength is:

The graph between the energy log $E$ of an electron and its de$-$Broglie wavelength log $\lambda$ will be

An electron beam,when accelerated by a voltage of $10 \ kV$,has a de-Broglie wavelength of $\lambda$. If the voltage is increased to $20 \ kV$,then the de-Broglie wavelength associated with the electron beam would be:

An alpha particle moves along a circular path of radius $0.5 \ mm$ in a magnetic field of $2 \times 10^{-2} \ T$. The de Broglie wavelength associated with the alpha particle is nearly (Planck's constant $= 6.63 \times 10^{-34} \ J \ s$)

$A$ proton and an electron are accelerated through the same potential difference. The ratio $\frac{\lambda_e}{\lambda_p}$ will be:

The de Broglie wavelength of a particle accelerated through a potential difference of $150 \ V$ is $10^{-10} \ m$. If it is accelerated through a potential difference of $600 \ V$,what will be its wavelength in $\mathring A$?