The resolving power of an electron microscope is greater than the resolving power of an optical microscope because

Assertion $(A):$ $A$ particle of mass $M$ at rest decays into two particles of masses $m_1$ and $m_2$,having non-zero velocities. The ratio of their de-Broglie wavelengths is unity.
Reason $(R):$ Here,we cannot apply the conservation of linear momentum.

$A$ beam of electrons of energy $E$ scatters from a target having atomic spacing of $1 \, Å$. The first maximum intensity occurs at $\theta = 60^{\circ}$. Then $E$ (in $eV$) is: (Planck constant $h = 6.64 \times 10^{-34} \, Js$, $1 \, eV = 1.6 \times 10^{-19} \, J$, electron mass $m = 9.1 \times 10^{-31} \, kg$)

If the de Broglie wavelength of an electron is equal to $10^{-3}$ times the wavelength of a photon of frequency $6 \times 10^{14} \, Hz,$ then the speed of the electron is equal to: (Speed of light $= 3 \times 10^8 \, m/s;$ Planck's constant $= 6.63 \times 10^{-34} \, J \cdot s;$ Mass of electron $= 9.1 \times 10^{-31} \, kg$)

Find the de Broglie wavelength associated with a Helium $(He)$ atom in Helium gas at room temperature $(27^{\circ}C)$ and $1$ atmospheric pressure.

What is the de-Broglie wavelength associated with an electron moving with a speed of $6.4 \times 10^{6} \ m/s$ (in $nm$)? $(m_{e} = 9.11 \times 10^{-31} \ kg, h = 6.63 \times 10^{-34} \ J \cdot s)$