MCQ
What will be de-Broglie wavelength of an electron moving with a velocity of $1.2 \times {10^5}\,m{s^{ - 1}}$
  • A
    $6.068 \times {10^{ - 9}}$
  • B
    $3.133 \times {10^{ - 37}}$
  • $6.626 \times {10^{ - 9}}$
  • D
    $6.018 \times {10^{ - 7}}$

Answer

Correct option: C.
$6.626 \times {10^{ - 9}}$
c
(c) $\lambda = \frac{h}{p}$, $p = mv$

$\lambda = \frac{h}{{mv}}$$ = \frac{{6.62 \times {{10}^{ - 34}}}}{{9.1 \times {{10}^{ - 31}} \times 1.2 \times {{10}^5}}}$

$\lambda = 6.626 \times {10^{ - 9}}m$.

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