The kinetic energy of an electron having de-Broglie wavelength $\lambda$ is ($h=$ Planck's constant,$m=$ mass of electron).

An electron is moving with initial velocity $\vec{v} = V_0 \hat{j}$ in a magnetic field $\vec{B} = B_0 \hat{i}$. Its de-Broglie wavelength will

The ratio of de Broglie wavelengths associated with thermal neutrons at temperatures $127^{\circ} C$ and $352^{\circ} C$ is

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?

An $\alpha$-particle,a proton,and an electron have the same kinetic energy. Which one of the following is correct in the case of their De-Broglie wavelength?

An electron of mass $m$ with an initial velocity $\vec{V} = V_0 \hat{i} \,(V_0 > 0)$ enters an electric field $\vec{E} = -E_0 \hat{i} \,(E_0 = \text{constant} > 0)$ at $t = 0$. If $\lambda_0$ is its de-Broglie wavelength initially,then its de-Broglie wavelength at time $t$ is: