The average life time of an electron in an excited state of hydrogen is 10-10 sec. How many revolutions does this electron make in the n = 2 state before dropping back to the ground state?
I have worked this exercise using classic physics but my answer is wrong. I've calculated the velocity of an electron in 2nd orbit, and I've obtained 1,09 E6, the radius of the second orbit, which results 2,12 E-10, and with that I can obtain time for one revolution, 1.21E-15, so by simply dividing 1E-10 by the time of one revolution I've obtained 8.23E4. However my answer is graded wrong. I have searched on internet and several books have my same answer, except one where for a time 1E-8 it proposed as result 4.03E06, but honestly I don't have any idea how to get that. All my alternatives go to the same result. Can anyone help me with the problem?
could it be a significant figure issue?
10^-10 is one sig fig ... so the answer can't have 3 sig fig
I'm not sure, maybe is a significant figure issue, but I only have one more attempt, so I would like to be sure. I have just found one example where a different result is proposed for a average lifetime of 1.0 E-8 (v = 0.6 x 10^6 m/sec; 4.73 x 10^6 revolutions)
But I don't know how it obtains these results. Maybe is something related with the velocity of the electron because here the author proposed 0.6E6 instead 1.09E06, but it doesn't have any sense, the velocity in the nth orbit is simply 2.182E6 divided by n, isn't it?
To calculate the number of revolutions an electron makes in the n = 2 state before dropping back to the ground state, we need to consider the concept of the Bohr model of the atom.
In the Bohr model, the energy levels of electrons in an atom are quantized, meaning they can only exist at certain discrete energy levels. When an electron absorbs energy and gets excited to a higher energy level, it eventually returns to its original ground state by emitting a photon.
The time it takes for an electron to remain in an excited state can be calculated using the uncertainty principle. The uncertainty principle states that the product of the uncertainty in energy and the uncertainty in time must be greater than or equal to a certain value.
In this case, the average lifetime of an electron in the n = 2 state is given as 10^-10 seconds. This means that the uncertainty in time is approximately 10^-10 seconds.
To find the number of revolutions, we need to calculate the time for one revolution in the n = 2 state. The orbital angular momentum of an electron in the n = 2 state is quantized and given by:
L = nħ
where L is the angular momentum, n is the principal quantum number (in this case, n = 2), and ħ is the reduced Planck's constant.
The angular momentum is related to the orbital velocity, v, and the radius of the orbit, r, by:
L = mvr
where m is the mass of the electron.
From classical physics, we can equate the centripetal force to the electrostatic force to find expressions for v and r:
mv^2/r = kZe^2/r^2
where k is Coulomb's constant, Z is the atomic number of hydrogen (which is 1), and e is the charge of the electron.
By solving these equations, we can find the velocity, v, and the radius, r, for an electron in the n = 2 state.
Once we have the velocity, we can calculate the time for one revolution, T, using the equation:
T = 2πr/v
This is the time it takes for the electron to complete one revolution in the n = 2 state.
Finally, to find the number of revolutions, we divide the average lifetime of an electron in the n = 2 state (10^-10 seconds) by the time for one revolution (T) computed above.
Nrev = (10^-10 seconds) / T
By plugging in the values for the mass of the electron, the reduced Planck's constant, Coulomb's constant, and the charge of the electron, and then performing the calculations, we can determine the correct answer to the problem.
Note: Quantum mechanics provides a more accurate description of the behavior of electrons in atoms, taking into account wave-particle duality and probability distributions. The Bohr model is a simplified approach that provides approximate answers in certain scenarios.