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Energy difference between ground state and excited state of an atom is 6.6 x 10^-34J^.The wave length of photon required to produce this transition is

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3 x 10⁸ cm

3 x10¹⁰ m

3 x 10^-8m

3 x 10⁸ m

answer is C.

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Detailed Solution

$λ = \frac{hc}{∆ E} = \frac{6.6 \times 10^{- 34} J - \sec \times 3 \times 10^{8} m \sec^{- 1}}{6.6 \times 10^{- 34} J} = 3 \times 10^{8} m$

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