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Question

# The energy of an electron in the second and the third Bohr orbit of hydrogen atom is −5.42×10−12 erg and −2.41×10−12 erg respectively. Calculate the wavelength (cm) of the emitted radiation when an electron drops from the third to the second orbit.

A
6.6×105
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B
6.6×107
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C
3×107
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D
3×105
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Solution

## The correct option is A 6.6×10−5We have, ΔE=E3−E2 ΔE=−2.41×10−12−(−5.42×10−12) erg ΔE=3.01×10−12 erg ΔE=hcλ λ=hcΔE λ=6.6×10−34 Js × 3×108 ms−13.01×10−12 erg λ=6.6×10−34 × 107erg s × 3×1010 cm s−13.01×10−12 erg λ=6.6×10−5 cm

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