Answer:
frequency transition 2.4626 10¹⁵ Hz
Step-by-step explanation:
Bohr's atomic model works very well for the hydrogen atom, the energy of the states is described by the expression
En = - 13.606 /n² eV
Where n is a positive integer
Let's calculate the energy for the base state n = 1 E1 = -13.606 eV
The energy for the first excited state n = 2 E2 = -13.6060 / 2²
E2 = -3.4015 Ev
The variation of the energy for the transition is
ΔE = E2 -E1
ΔE= -3.4015 +13.606
ΔE = 10.2045 eV
We use the Planck equation to find the frequency
E = 10.2045 eV (1.6 10⁻¹⁹ J/1 eV) = 16.3272 10⁻¹⁹ J
E = h f
f = E / h
f = 16.3272 10⁻¹⁹ / 6.63 10⁻³⁴
f = 2.4626 10¹⁵ Hz
Let's calculate the frequency of the excited state
E2 = 3.4015 eV (1.6 10⁻¹⁹ J/1eV) = 5.4424 10⁻¹⁹ J
f = 5.4424 10⁻¹⁹ / 6.63 10⁻³⁴
f = 0.8209 10¹⁵ Hz
To make the comparison, divide the two frequencies
f transition / f excited = 2.4626 / 0.8209
f transition / f excited = 3
The frequency of the transition is 3 times greater than the frequency of the first excited state