Question

At 200 degree celsius,the velocity of hydrogen molecule is 2×10 power 5 cm/sec . in this case , the debroglie wavelength is about​

Answer 1

mass of one mole of hydrogen molecules = 2g

we know, 1 mole of hydrogen molecules = 6.022 × 10²³ hydrogen molecules

so, mass of each hydrogen molecule, m = 2g/6.022 × 10²³ = 0.3321 × 10-²³ g

= 3.321 × 10^-24 g

= 3.321 × 10^-27 kg [ as we know, 1g = 10^-3 kg ]

given, velocity of hydrogen molecule , v = 2 × 10^5 cm/s = 2 × 10³ m/s

using De-broglie's wavelength equation,

wavelength = h/mv

h is Plank's constant i.e., h = 6.63 × 10^-34 J.s

so, wavelength = 6.63 × 10^-34/(3.321 × 10^-27 × 2 × 10³)

= (6.63/3.321 × 2) × 10^(-34 + 24)

= 0.9981 × 10^-10 m

= 0.9981 A°

hence, De-broglie's wavelength is about 0.9981 A° ≈ 1 A°

Answer 2

Hey Dear,

◆ Answer -

λ = 1 A° ...(aprox)

● Explanation -

# Given -

v = 2×10^5 cm/s = 2×10^3 m/s

# Solution -

Mass of hydrogen molecule can be calculated as -

Mass of hydrogen molecule = molecular mass / Avogadro's number

m = 2×10^-3 / 6.022×10^23

m = 2×10^-26 / 6.022

m = 3.32×10^-27 kg

de Broglie's wavelength of hydrogen atom is calculated as -

λ = h / mv

Substituting values,

λ = 6.626×10^-34 / (3.32×10^-27 × 2×10^3)

λ = 9.98×10^-11 m

λ = 1 A° ...(aprox)

Therefore, de Broglie's wavelength of hydrogen will be approximately 1 A°.

Thanks dear...