Question

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

Answer 1

$<b>$
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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

→ [B]


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Answer 2

Here is your answer ⤵⤵⤵

(b)

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