An electron is moving with an initial velocity $\vec{V} = V_{0} \hat{i}$ and is in a uniform magnetic field $\vec{B} = B_{0} \hat{j}$. Then its de Broglie wavelength

The de-Broglie wavelength $\lambda$ of a particle is:

What is the theoretical formula for the de-Broglie wavelength of a particle?

The graph which shows the variation of the de Broglie wavelength $(\lambda)$ of a particle and its associated momentum $(p)$ is

The de Broglie wavelength of an oxygen molecule at $27^{\circ} C$ is $x \times 10^{-12} \ m$. The value of $x$ is (take Planck's constant $= 6.63 \times 10^{-34} \ J \cdot s$, Boltzmann constant $= 1.38 \times 10^{-23} \ J/K$, mass of oxygen molecule $= 5.31 \times 10^{-26} \ kg$).

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 $....\, nm$.
$[m_e = \text{mass of electron} = 9 \times 10^{-31}\, kg, h = \text{Planck constant} = 6.6 \times 10^{-34}\, Js, k_B = \text{Boltzmann constant} = 1.38 \times 10^{-23}\, JK^{-1}]$