Question

The velocity of helium atom at 300 K is $2.40\times {10}^2$ meter per sec. What is its wavelength?( mass number of helium is 4)

• A. $0.414nm$
• B. $0.83nm$
• C. $803\mathring{A}$
• D. $800\mathring{A}$

Answer 1

The correct option is A $0.414nm$

Average velocity of a helium atom is $2.4\times 10^2$ m/s=240 m/s

Mass of one helium atom,m=$46.023\times 10^{23}$g

$=4\times 1.66\times 10^{-27}kg$

$=6.64\times 10^{-27}kg$

By using De-broglie wavelength:

$\lambda =\frac{h}{mv}$ (here h is plank's constant)

$\lambda =\frac{6.6\times {10}^{-34}}{6.64\times {10}^{-27}\times 240}$

$=0.004141\times 10⁻⁷m$

$=4.141\times {10}^{-10}m$

$1m={10}^{-9}nm$

$=4.141\times {10}^{-10}m$ = $0.414$ nm

So, the correct option is$A$