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The ionization energy of a hydrogen-like Bohr atom is 4 Rdybergs. Find the wavelength of radiation emitted when the electron jumps from the first excited state to the ground state:
$[1 R y d b e r g = 2.2 \times 10^{- 18} j o u l e]$
$[h = 6.6 \times 10^{- 34} J s, c = 3 \times 10^{8} m / s .]$
Bohr radius of hydrogen atom $= 5 \times 10^{- 11} m]$

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600 $\overset{°}{A}$

500 $\overset{°}{A}$

400 $\overset{°}{A}$

300 $\overset{°}{A}$

answer is B.

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Detailed Solution

The energy in the ground state

$E_{1} = - 4$ Rydberg

$E_{1} = - 4 \times 2.2 \times 10^{18} J$

The energy of the first excited state (n=2)

$E_{2} = \frac{E_{1}}{4} = - 2.2 \times 10^{- 18} J$

The energy difference

$Δ E = E_{2} - E_{1} = 3 \times 2.2 \times 10^{- 18} J$

Now, the wavelength of radiation emitted is

$λ = \frac{h c}{Δ E}$

$λ = \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{3 \times 2.2 \times 10^{- 18}} = 300 \overset{\circ}{A}$

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