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The deBroglie wavelength of electron in ground state of hydrogen atom is (given that the radius of the first orbit of hydrogen atom’s 0.53 Å) (1) 0.53 Å (2) 3.33 Å (3) 1.67 Å (4) 3.33×10–6 m |
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Answer» Correct option : (2) 3.33 Å Explanation: Since the electron is in ground state which implies that electron is in the first orbit of hydrogen atom with radius 0.53 Å. For first orbit the circumference of the orbit 2πr is the deBroglie wavelength. Hence λ = 2πr = 2 x 0.53 x π = 3.33 Å |
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