Question

The average speed of a helium atom at ${25}^{o}C$ is $1.23\times {10}^{3}m{s}^{-1}$. What is the average wavelength of helium atom at ${25}^{o}C$?

• A. $1.35\times {10}^{-31}$
• B. $1.35\times {10}^{-34}$
• C. $1.35\times {10}^{-30}$
• D. $1.35\times {10}^{-32}$

Answer 1

Answer: (B)

$1.35\times {10}^{-34}$

According to de-Broglie's principle,
$\lambda =h/mv$ so
${\lambda }_{av}=\frac{h}{m{u}_{av}}=\frac{6.626\times {10}^{-34}}{4\times {10}^{-3}\times 1.23\times {10}^3}$
$=1.35\times {10}^{-34}$