Question

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

• A. 2 $\mathring{A} $
• B. 4 $\mathring{A} $
• C. 8 $\mathring{A} $
• D. 100 $\mathring{A} $

Answer 1

Answer: (B)

According to de-Broglie
$\lambda=\frac{h}{mv}=\frac{6.62\times10^{-20} erg. sec}{\frac{2}{6.023 \times10^{23}}\times5\times10^{4}cm/sec } $
$=\frac{6.62\times10^{-27}\times6.023 \times10^{23}}{2\times5\times10^{4}}cm$
$=4\times10^{-8} cm =4 \,\mathring{A} $