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For a hydrogen atom, calculate the wavelength of light (in m) that

would be emitted for the orbital transition of n = 5 to n = 1. Submit an
answer to four significant figures.

User Bublitz
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1 Answer

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Answer:

Step-by-step explanation:
The energy released when an electron transitions from energy level n1 to energy level n2 in a hydrogen atom is given by the equation:

ΔE = E2 - E1 = (-2.18 x 10^-18 J)(1/n1^2 - 1/n2^2)

where E1 and E2 are the energies of the initial and final states, respectively.

For the transition from n=5 to n=1 in a hydrogen atom, we have:

ΔE = (-2.18 x 10^-18 J)(1/1^2 - 1/5^2) = -1.7104 x 10^-18 J

The energy of a photon of light is given by the equation:

E = hc/λ

where h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength of the light.

Solving for λ, we get:

λ = hc/ΔE = (6.626 x 10^-34 J s)(2.998 x 10^8 m/s)/(-1.7104 x 10^-18 J) ≈ 1.215 x 10^-7 m

Rounding to four significant figures, the wavelength of light emitted in this transition is approximately 1.215 x 10^-7 m.

User Andrew Rahimov
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