could it be a significant figure issue?
10^-10 is one sig fig ... so the answer can't have 3 sig fig
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?
2 answers
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?
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?