If the radius of the first orbit of the hydrogen atom is 0.53 A∘ , then the de-Broglie wavelength of the electron in the ground state of hydrogen atom…

Question

If the radius of the first orbit of the hydrogen atom is 0.53 A∘ , then the de-Broglie wavelength of the electron in the ground state of hydrogen atom will be

MARKS Content Team · Accepted Answer

Correct answer: (B) 3.33 A∘. According to Bohr's quantization condition mvr=nh/(2π) ∴ (h )/mv= (2πr)/n ... (i) ∵ de-Broglie wavelength λ=h/mv ... (ii) From Eqs. (i) and (ii), λ=(2πr)/n ∵ Electron in the ground state r=0.53 A∘ and n = 1 ⇒λ=(2π×0.53 A∘)/1 =3.33 A∘

If the radius of the first orbit of the hydrogen atom is $0.53 \overset{\circ}{A}$ , then the de-Broglie wavelength of the electron in the ground state of hydrogen atom will be

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