If the kinetic energy of an electron is increased four times,the wavelength of the de-Broglie wave associated with it would become

The kinetic energy of an electron gets tripled,then the de-Broglie wavelength associated with it changes by a factor of:

What is the $(a)$ momentum,$(b)$ speed,and $(c)$ de Broglie wavelength of an electron with kinetic energy of $120 \ eV$?

$A$ proton and an $\alpha$-particle are accelerated through the same potential difference $V$. The ratio of their de-Broglie wavelengths is

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength $\lambda$. When the potential is changed to $V_2$,its de-Broglie wavelength increases by $50 \%$. The value of $\left(\frac{V_1}{V_2}\right)$ is

The de Broglie wavelength of an electron in a metal at $27^{\circ}C$ is compared with the given mean distance between two electrons in the metal,which is $2 \times 10^{-10} \ m$. The ratio of the mean distance to the de Broglie wavelength is approximately: