If the de-Broglie wavelength at a temperature of $27^oC$ is $\lambda$,what will be the de-Broglie wavelength at $927^oC$?

Crystal diffraction experiments can be performed using $X-$rays,or electrons accelerated through an appropriate voltage. Which probe has greater energy? (For quantitative comparison,take the wavelength of the probe equal to $1\,\mathring{A}$,which is of the order of inter-atomic spacing in the lattice) $(m_{e}=9.11 \times 10^{-31}\,kg)$

The de Broglie wavelength of an electron moving with kinetic energy of $144 \;eV$ is nearly

The kinetic energy of an electron is $5 \ eV$. Calculate the de-Broglie wavelength associated with it in $\mathring{A}$. $(h = 6.6 \times 10^{-34} \ J \cdot s, m_e = 9.1 \times 10^{-31} \ kg)$

$A$ photosensitive metallic surface emits electrons when $X$-rays of wavelength $\lambda$ fall on it. The de-Broglie wavelength of the emitted electrons is (Neglect the work function of the surface,$m$ is mass of the electron,$h$ is Planck's constant,$c$ is the velocity of light).

If the potential difference used to accelerate electrons is doubled,by what factor does the de-Broglie wavelength associated with the electrons change?