What is the de-Broglie wavelength of a bullet of mass $0.033$ kg travelling at the speed of $1$ km/s? $(h = 6.6 \times 10^{-34} \text{ Js})$

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. 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 equal to :

What is the de Broglie wavelength of
$(a)$ a bullet of mass $0.040 \; kg$ travelling at the speed of $1.0 \; km/s$
$(b)$ a ball of mass $0.060 \; kg$ moving at a speed of $1.0 \; m/s$,and
$(c)$ a dust particle of mass $1.0 \times 10^{-9} \; kg$ drifting with a speed of $2.2 \; m/s$?

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?

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

Find the ratio of the de Broglie wavelengths of a proton and an $\alpha$-particle accelerated through the same potential difference.