velocity of helium atom at 300 K is 2.40*10^2 meter per sec. what is its wavelength

Dear Student

Velocity of Helium Atom, v = 2.40 x 10² m/sec

Wavelength = ?

we know, 

Mass number of He = 4

Also, p = $\frac{h}{\lambda }$
So, λ = $\frac{h}{p}$  [where, h is planck's constant = 6.626 x 10^-34 m² kg/sec, p = momentum and λ is the wavelength]

because, p = mass x velocity

Therefore, 

λ​ = $\frac{h}{m x v}$ = $\frac{6.626 x {10}^{-34} m^2 kg/se{c}^{ }}{m x 2.40 x {10}^2 m/sec}$

Here, m is the mass of helium atom

So, mass of a single atom of He in kg will be calculated as:

Mass number of He = 4

Thus, m = $\frac{4}{1000}\times \frac{1}{Avagadro number}$ = $\frac{4}{1000}x \frac{1}{6.023 x {10}^{23}}$ = 6.647 x 10^-27 kg


Hence, 

λ​ = ​$\frac{6.626 x {10}^{-34} m^2 kg/se{c}^{ }}{6.647 x {10}^{-27} kg x 2.40 x {10}^2 m/sec}$ = $\frac{6.626 x {{10}^{-}}^{34} m}{15.9528 x {10}^{-25}}$

  = 0.41535 x 10^{-34 + 25} = 0.41535 x 10^-9 m = 4.1535 x 10^-8 m

Since, 1 m = 10⁹ nm

Hence, this makes ​λ = 0.41535 x 10^{-9 + 9} nm = 0.41535 nm or 0.416 nm
So ans is A

​Regards