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Question

The energy of an electron in the second and the third Bohr's orbits of the hydrogen atom is 5.42×1012 erg and 2.41×1012 erg respectively. Calculate the wavelength of the emitted radiation when the electron drops from the third orbit to the second orbit.


Correct Answer
A
6.604×103A
Your Answer
B
6.604A
Your Answer
C
3.905×103A
Your Answer
D
3.905A

Solution

The correct option is A 6.604×103A
Energy 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=E3E2=hcλ
λ=hcE3E2
Given E2=5.42×1012erg and E3=2.41×1012erg
We know that c=3×1010cm/s.
λ=6.626 × 1027 ×3 × 10102.41 ×1012(5.42 × 1012)= 19.878×10173.01×1012=6.604×105cm=6.604×103A

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