Question

При переходе электрона в атоме водорода из одного стационарного состояния в другое излучаются фотоны с энергией 3,03 х10-19 Дж. Определите частоту излучения атома.

Answer 1

Е=hv
v=E/h
v=3.03*10-19/6.62*10-34

Answer 2

Calculation of the Frequency of Radiation

To determine the frequency of the radiation emitted by the hydrogen atom when an electron transitions from one stationary state to another, we can use the equation:

E = hf

where: - E is the energy of the photon (given as 3.03 x 10^-19 J) - h is Planck's constant (6.626 x 10^-34 J·s) - f is the frequency of the radiation

To find the frequency, we rearrange the equation:

f = E / h

Substituting the given values:

f = (3.03 x 10^-19 J) / (6.626 x 10^-34 J·s)

Calculating this expression gives us the frequency of the radiation emitted by the hydrogen atom.

Let's calculate it.