Question

При переходе электрона в атоме водорода с третьей стационарной орбиты на вторую излучается фотон. Если при излучении фотона атом теряет энергию 1,9 эВ, то частота этого излучения?????

Answer 1

Дано Е=1,9 эВ    v- ?


v=E\h= 1,9*1,6*10^-19\6,63*10^-34=0.458 *10^15 Гц

Answer 2

Calculation of the Frequency of the Emitted Photon

To calculate the frequency of the emitted photon when an electron transitions from the third stationary orbit to the second orbit in a hydrogen atom, we can use the equation:

E = hf

where: - E is the energy of the photon, - h is Planck's constant (approximately 6.626 x 10^-34 J·s), - f is the frequency of the photon.

Given that the atom loses 1.9 eV of energy during the emission of the photon, we need to convert this energy to joules before using the equation.

1 eV is equal to 1.602 x 10^-19 J. Therefore, 1.9 eV is equal to 1.9 x 1.602 x 10^-19 J.

Now, we can substitute the energy value into the equation and solve for the frequency:

1.9 x 1.602 x 10^-19 J = (6.626 x 10^-34 J·s) x f

Simplifying the equation:

f = (1.9 x 1.602 x 10^-19 J) / (6.626 x 10^-34 J·s)

Calculating the frequency using the given values:

f ≈ 4.57 x 10^14 Hz

Therefore, the frequency of the emitted photon is approximately 4.57 x 10^14 Hz.