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[Solved] The de-Broglie wavelength of an electron in the ground state of hydrogen atom is :

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The de-Broglie wavelength of an electron in the ground state of hydrogen atom is :

[K.E. = 13.6 eV ; 1eV = 1.602 x 10-19 J]

(a) 33.28 nm

(b) 3.328 nm

(c) 0.3328 nm

(d) 0.0332 nm

1 Answer
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(c) 0.3328 nm

Explanation:

We know that

K.E = 1/2 mv2

= 0.3328 × 10-9 = 0.3328 nm
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