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JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

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

\left({K} . {E} .=13.6 {eV} ; 1 {eV}=1.602 \times 10^{-19} {~J}\right)

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Correct option: C

0.3328 {~nm}

Option C (0.3328 nm) is correct because it is calculated using the de-Broglie wavelength formula with the given kinetic energy.

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