The electron in the hydrogen atom moves to a second stationary orbit with an energy of E_(2)=-3.4 eV, emitting waves whose length is 434 nm.Find the number of the orbit in which this electron was located before moving to the second stationary orbit. Planck's constant h=6.63ast 10^-34 Jxs, elementary charge e=1.6^ast 10^-19C The speed of light in vacuum is c=3^ast 10^8m/s Give your answer in whole numbers. Youranswer

To find the number of the orbit in which the electron was located before moving to the second stationary orbit, we can use the formula for the energy levels of a hydrogen atom:<br /><br />$E_n = -\frac{13.6 \text{ eV}}{n^2}$<br /><br />where $E_n$ is the energy of the electron in the nth orbit, and $n$ is the principal quantum number (the number of the orbit).<br /><br />We know that the energy of the electron in the second stationary orbit is $E_2 = -3.4$ eV. We can use this information to find the principal quantum number of the orbit from which the electron transitioned to the second orbit.<br /><br />The energy emitted during the transition is given by the difference in energy levels:<br /><br />$\Delta E = E_2 - E_1$<br /><br />where $E_1$ is the energy of the electron in the first orbit. We can calculate $E_1$ using the formula for the energy levels of a hydrogen atom:<br /><br />$E_1 = -\frac{13.6 \text{ eV}}{1^2} = -13.6$ eV<br /><br />Now we can calculate $\Delta E$:<br /><br />$\Delta E = E_2 - E_1 = -3.4 \text{ eV} - (-13.6 \text{ eV}) = 10.2 \text{ eV}$<br /><br />The energy of the emitted wave is related to its wavelength by the equation:<br /><br />$E = \frac{hc}{\lambda}$<br /><br />where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength of the emitted wave. We can rearrange this equation to solve for the energy:<br /><br />$E = \frac{6.63 \times 10^{-34} \text{ J s} \times 3 \times 10^8 \text{ m/s}}{434 \times 10^{-9} \text{ m}} = 4.59 \times 10^{-19} \text{ J}$<br /><br />Since $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$, we can convert the energy to eV:<br /><br />$E = \frac{4.59 \times 10^{-19} \text{ J}}{1.6 \times 10^{-19} \text{ J/eV}} = 2.87 \text{ eV}$<br /><br />Now we can use the energy difference to find the principal quantum number of the initial orbit:<br /><br />$\Delta E = E_2 - E_1 = -\frac{13.6 \text{ eV}}{n_1^2} - (-13.6 \text{ eV}) = \frac{13.6 \text{ eV}}{n_1^2}$<br /><br />Solving for $n_1$, we get:<br /><br />$n_1 = \sqrt{\frac{13.6 \text{ eV}}{\Delta E}} = \sqrt{\frac{13.6 \text{ eV}}{2.87 \text{ eV}}} = 3$<br /><br />Therefore, the number of the orbit in which the electron was located before moving to the second stationary orbit is 3.