If an electron in $n = 3$ orbit of hydrogen atom jumps down to $n = 2$ orbit, the amount of energy released and the wavelength of radiation emitted are

0.85 e V, 6566 ˚ A

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1.89 e V, 1240 ˚ A

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1.89 e V, 6566 ˚ A

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1.5 e V, 6566 ˚ A

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Solution

The correct option is C $1.89 e V, 6566 ˚ A$
$E_{3 ⟶ 2} = E_{3} - E_{2} E_{3} = - \frac{13.6}{9} = - 1.51 e V E_{2} = - \frac{13.6}{4} = - 3.4 e V$
Therefore, energy released is $= 3.4 - 1.51 = 1.89 e V$
$\frac{1}{λ} = (1.097 \times 10^{7}) (\frac{1}{4} - \frac{1}{9}) ∴ λ = 6560 \times 10^{- 10} m$
Therefore, wavelength emitted is, $λ = 6560 ˚ A$

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If an electron in n=3 orbit of hydrogen atom jumps down to n=2 orbit, the amount of energy released and the wavelength of radiation emitted are

The correct option is C 1.89 eV,6566˚AE3⟶2=E3−E2E3=−13.69=−1.51eVE2=−13.64=−3.4eVTherefore, energy released is =3.4−1.51=1.89 eV 1λ=(1.097×107)(14−19)∴ λ=6560×10−10mTherefore, wavelength emitted is, λ=6560˚A

If an electron in $n = 3$ orbit of hydrogen atom jumps down to $n = 2$ orbit, the amount of energy released and the wavelength of radiation emitted are