Complete question is;
What is the frequency of light emitted when the electron in a hydrogen atom undergoes a transition from energy level n=6 to level n=3?
Answer:
Frequency = 2.742 × 10^(14) s^(-1)
Explanation:
First of all, the energy of hydrogen electron from online values is;
E_n = -2.18 × 10^(-18) × (1/n²) J
n is the principal quantum number
We are told that hydrogen atom undergoes a transition from energy levels n = 3 to n = 6.
Thus, it means we have to find the difference between the electrons energy in the energy levels n = 3 to n = 6.
Thus;
E_n = E_6 - E_3
Thus;
E_n = [-2.18 × 10^(-18) × (1/6²)] - [-2.18 × 10^(-18) × (1/3²)]
E_n = (2.18 × 10^(-18)) × [-1/36 + 1/9]
E_n = 0.1817 × 10^(-18) J
From Planck expression, we can find the frequency. Thus;
E = hf
Where h is Planck's constant = 6.626 × 10^(-34) m²kg/s
Thus;
0.1817 × 10^(-18) = 6.626 × 10^(-34) × f
f = (0.1817 × 10^(-18))/(6.626 × 10^(-34))
f = 2.742 × 10^(14) s^(-1)
