If the de-Broglie wavelengths for a proton and for an $\alpha$-particle are equal, then the ratio of their velocities will be

The de-Broglie wavelength of a moving bus with speed $v$ is $\lambda$. Some passengers left the bus at a stoppage. Now, when the bus moves with twice its initial speed, its kinetic energy is found to be twice its initial value. What is the de-Broglie wavelength of the bus now?

The equation $\lambda = \frac{1.227}{x} \text{ nm}$ can be used to find the de-Broglie wavelength of an electron. In this equation,$x$ stands for:

How much energy is imparted to an electron so that its de-Broglie wavelength reduces from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$? (Let $E$ be the initial energy of the electron).

The temperature of an ideal gas in $3$-dimensions is $300\, K$. The corresponding de-Broglie wavelength of the electron approximately at $300\, K$ is $....\, nm$.
$[m_e = \text{mass of electron} = 9 \times 10^{-31}\, kg, h = \text{Planck constant} = 6.6 \times 10^{-34}\, Js, k_B = \text{Boltzmann constant} = 1.38 \times 10^{-23}\, JK^{-1}]$

The de Broglie wavelength of an electron accelerated to a potential of $ 400 \,V $ is approximately (in $\,nm$)