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Question

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

A
6.604×103A
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B
6.604A
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C
3.905×103A
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D
3.905A
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Solution

## The correct option is A 6.604×103∘AEnergy of an electron in the nth orbit is denoted as En. So, the energies of orbits 3 and 2 can be written as E3 and E2 respectively. ΔE is the difference between these energies. If λ is the wavelength of the emitted radiation, we can say that: ΔE=E3−E2=hcλ ⟹λ=hcE3−E2 Given E2=−5.42×10−12erg and E3=−2.41×10−12erg We know that c=3×1010cm/s. ∴λ=6.626 × 10−27 ×3 × 1010−2.41 ×10−12−(−5.42 × 10−12)= 19.878×10−173.01×10−12=6.604×10−5cm=6.604×103∘A

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