Q 12-11-019NEETJEE MainEAMCET 2001 (Engineering)Easy
The de Broglie wavelength of an electron having $80$ eV of energy is nearly ($1$ eV $= 1.6 \times 10^{-19}$ J, mass of electron $= 9 \times 10^{-31}$ kg, Planck's constant $= 6.6 \times 10^{-34}$ J s)
Answer: (D) $1.4$ Å
Concept. $\lambda = \dfrac{h}{\sqrt{2mE}}$.
Step 1. Substitute.
$$\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9 \times 10^{-31} \times 80 \times 1.6 \times 10^{-19}}} = \frac{6.6 \times 10^{-34}}{4.8 \times 10^{-24}} \approx 1.4\ \text{Å}$$
Answer. $1.4$ Å (Option D).
Solution by Sreeraj P, M.Sc Physics