Question

What is ΔE (change in energy) for the transition of an electron from n = 7 to n = 5 in a hydrogen atom? What is the frequency, ν, of the spectral line produced?

A. ΔE = -2.19E-19 J, ν = 4.57E14 s^-1
B. ΔE = 2.19E-19 J, ν = 4.57E14 s^-1
C. ΔE = -3.41E-19 J, ν = 5.09E14 s^-1
D. ΔE = 3.41E-19 J, ν = 5.09E14 s^-1

Answer 1

The change in energy, ΔE, for the transition of an electron from n = 7 to n = 5 in a hydrogen atom is 1.9 eV. The frequency, ν, of the spectral line produced is approximately 2.26 × 10^15 s^-1.

Step-by-step explanation:

The change in energy, ΔE, for the transition of an electron from n = 7 to n = 5 in a hydrogen atom can be calculated using the formula:

ΔE = En=5 - En=7

Given that E2 = -3.4 eV and E5 = -1.5 eV, we can substitute these values into the formula to find:

ΔE = -1.5 eV - (-3.4 eV) = 1.9 eV

To calculate the frequency, ν, of the spectral line produced, we can use the equation:

ΔE = hν, where h is Planck's constant (6.626 × 10-34 J·s).

Substituting the calculated value of ΔE and Planck's constant:

1.9 eV = (6.626 × 10-34 J·s)ν

Solving for ν, we find:

ν ≈ 2.26 × 1015 s-1

Therefore, the correct answer is D. ΔE = 3.41E-19 J, ν = 5.09E14 s^-1.