Question

If the velocity of hydrogen molecule is $5\times 10^4$ cm $se{c}^{-1}$, 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
$\lambda =\frac{h}{mv}=\frac{6.62\times 10^{-20}erg.sec}{\frac{2}{6.023\times 10^{23}}\times 5\times 10^{4}cm/sec}$
$=\frac{6.62\times 10^{-27}\times 6.023\times 10^{23}}{2\times 5\times 10^4}cm$
$=4\times 10^{-8}cm=4\mathring{A}$