Q61 / 90 · JEE Main 2021 (01 Sep Shift 2) · Physics

The temperature of an ideal gas in three dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K is:

me=mass of electron =9×10-31 kg, h=Planck constant =6.6×10-34 J s, kB=Boltzmann constant=1.38×10-23 J K-1

  • A
    2.26 nm
  • B
    3.25 nm
  • C
    8.46 nm
  • Correct answer
    D
    6.26 nm

Solution

k=1.38×10-23 J K-1

Mass of electron =9×10-31 kg

⇒λD=h2m×32×kT=h3mkT

⇒λD=6.6×10-343×9×10-31×1.38×10-23×300=6.6×10-34105.72×10-27=62 Aο

Thus x=62 Aο

Hence, the correct answer is option: 'D'.

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