MCQ
De – Broglie wavelength of an electron accelerating by $100 V$
- ✓$1.227 Å$
- B$122.7 μm$
- C$0.2227 m$
- D$12.27 Å$
✓
Answer
Correct option: A.
$1.227 Å$
a) $1.227 Å$
Solution:
$\lambda=\frac{12.27}{\sqrt{V}}$
$=\frac{12.27}{\sqrt{100}} $
$=\frac{12.27}{10}$
$\lambda=1.227 Å$
Solution:
$\lambda=\frac{12.27}{\sqrt{V}}$
$=\frac{12.27}{\sqrt{100}} $
$=\frac{12.27}{10}$
$\lambda=1.227 Å$
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