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Q.

The de Broglie wavelength of the electron in the ground state of hydrogen atom is  [K.E=13.6eV]; 1eV=1.602×1019J 

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a

0.3328 nm

b

3.325 nm

c

33.28 nm

d

0.0332 nm 

answer is C.

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Detailed Solution

λ=h2mKE=6.626×1034JS2×9.1×1031×1.602×1019×13.6

=0.3328nm

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The de Broglie wavelength of the electron in the ground state of hydrogen atom is  [K.E=13.6eV];  1eV=1.602×10−19J