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Q.

The de Broglie wavelength of the electron in the ground state of hydrogen atom is $[K . E = 13.6 e V];$ $1 e V = 1.602 \times 10^{- 19} J$

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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

$λ = \frac{h}{\sqrt{2 m K E}} = \frac{6.626 \times 10^{- 34} J S}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 1.602 \times 10^{- 19} \times 13.6}}$

$= 0.3328 n m$

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