Question

The de-Broglie wavelength of an electron having $80eV$ of energy is nearly

(1eV = 1.6 × 10^–19 J, Mass of electron = 9 × 10^–31kg

Plank’s constant = 6.6 × 10^–34 J-sec)

• A. 140 Å
• B. 0.14 Å
• C. 14 Å
• D. 1.4 Å

Answer 1

Answer: (D)

1.4 Å

Answer 2

$\lambda = \frac{h}{\sqrt{2mE}} = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9 \times 10^{- 31} \times 80 \times 1.6 \times 10^{- 19}}} = 1.4Å$