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Question: If the velocity of hydrogen molecule is \(5 \times 10^{4}cmsec^{- 1}\), then its de-Broglie waveleng...

If the velocity of hydrogen molecule is 5×104cmsec−15 \times 10^{4}cmsec^{- 1}, then its de-Broglie wavelength is

A

2 Å

B

4 Å

C

8 Å

D

100 Å

Answer

4 Å

Explanation

Solution

According to de–Broglie

λ=hmv=6.62×10−20erg.sec⁡26.023×1023×5×104cm/sec⁡\lambda = \frac{h}{mv} = \frac{6.62 \times 10^{- 20}erg.\sec}{\frac{2}{6.023 \times 10^{23}} \times 5 \times 10^{4}cm/\sec}

=6.62×10−27×6.023×10232×5×104cm=4×10−8cm=4A˚= \frac{6.62 \times 10^{- 27} \times 6.023 \times 10^{23}}{2 \times 5 \times 10^{4}}cm = 4 \times 10^{- 8}cm = 4Å.