If the velocity of hydrogen molecule is 5 × 104 cm sec−1, then its de-Broglie wavelength is

A
$2 A^{o}$
B
$4 A^{o}$
C
$8 A^{o}$
D
$100 A^{o}$

Solution:

$\begin{array}{l} λ = \frac{h}{mv} = \frac{6 .626 \times 10^{- 27} erg . \sec}{\frac{2}{6 .023 \times 10^{23}} \times 5 \times 10^{4} cm / \sec} \\ Since mass of a Hydrogen molecule = \frac{2}{6.023 \times 10^{23}} \\ = \frac{6 .62 \times 10^{- 27} \times 6 .023 \times 10^{23}}{2 \times 5 \times 10^{4}} cm \\ = 4 \times 10^{- 8} cm = 4 A^{o} . \end{array}$

If the velocity of hydrogen molecule is $5 \times 10^{4} cm \sec^{- 1},$ then its de-Broglie wavelength is

Solution:

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