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

If the potential difference used to accelerate electrons is doubled,by what factor does the de Broglie wavelength $(\lambda)$ associated with the electrons change?

An electron,an $\alpha$-particle,and a proton have the same kinetic energy. Which of these particles has the shortest de Broglie wavelength?

Assertion $(A):$ $A$ particle of mass $M$ at rest decays into two particles of masses $m_1$ and $m_2$,having non-zero velocities. The ratio of their de-Broglie wavelengths is unity.
Reason $(R):$ Here,we cannot apply the conservation of linear momentum.

The wavelength associated with an electron accelerated through a potential difference of $100 \ V$ is nearly .............. $ \mathring A $

The de Broglie wavelengths of an electron $(e)$,proton $(p)$,neutron $(n)$,and $\alpha$-particle are $\lambda_e, \lambda_p, \lambda_n$,and $\lambda_\alpha$ respectively. All have the same kinetic energy of $1 \ MeV$. Which of the following represents the correct increasing order of their wavelengths?