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

The de-Broglie wavelength of an electron is $0.4 \times 10^{-10} \ m$ when its kinetic energy is $1.0 \ keV$. Its wavelength will be $1.0 \times 10^{-10} \ m$ when its kinetic energy is: (in $keV$)

The de-Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of $3 \ km/s$ will be ............ $\mathring{A}$.

The potential energy of a particle of mass $m$ is given by $U(x) = \begin{cases} E_0; & 0 \le x \le 1 \\ 0; & x > 1 \end{cases}$. $\lambda_1$ and $\lambda_2$ are the de-Broglie wavelengths of the particle when $0 \le x \le 1$ and $x > 1$ respectively. If the total energy of the particle is $2 E_0$,the ratio $\frac{\lambda_1}{\lambda_2}$ will be:

An electron microscope uses electrons of $40 \ keV$. The de Broglie wavelength associated with these electrons is approximately:

The de Broglie wavelength of a neutron at $27^{\circ}C$ is $\lambda$. What will be its de Broglie wavelength at $927^{\circ}C$?