Question

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

• A. 2 Å
• B. 4 Å
• C. 8 Å
• D. 100 Å

Answer 1

The correct option is

D

4 Å

According to de-Broglie

$\lambda =\frac{h}{mv}=\frac{6.62\times {10}^{-20}erg.sec}{\frac{2}{6.023\times {10}^{23}}\times 5\times {10}^{4}cm/sec}$

= $\frac{6.62\times {10}^{-27}\times 6.023\times {10}^{23}}{2\times 5\times {10}^4}cm$

= $4\times {10}^{-8}cm=4\AA$

As per rules of division and multiplication 2 has to be multiplied and the Avogadro number moves up.