Energy of an electron in an excited hydrogen atom is – 3.4 eV. Its angular momentum will be: $h = 6.626 \times 10^{- 34} J - s$

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$1.51 \times 10^{- 31} J \sec$

$3.72 \times 10^{- 34} J \sec$

$2.11 \times 10^{- 34} J \sec$

$1.11 \times 10^{34} J \sec$

answer is C.

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Detailed Solution

The electron is in the second orbit (n=2)
Hence $L = \frac{nh}{2 π} = \frac{2 h}{2 π} = \frac{6.6 \times 10^{- 34}}{π}$ $= 2.11 \times 10^{- 34} J - \sec$

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