Please show work and explain. Answer is given already. Thank You. An electron in
ID: 1302767 • Letter: P
Question
Please show work and explain. Answer is given already. Thank You. An electron in an excited hydrogen atom makes two transitions. First the electron drops from the n 5 to the n 2 state, then the electron drops from the n 2 to the n i state. compare the energy and wavelength of the photon emitted in each transition. a) in the 1st transition the energy of the photon and the wavelength of the photon are both greater b in the 1sttransition the energy of the photon is greater and the wavelength of the photon is less gnin the 1st transition the energy of the photon is less and the wavelength of the photon is greater d in the 1" transition the energy of the photon and the wavelength of the photon are both less g) none of the above 2) Calculate the frequency of the photon emitted in the first transition 3) Calculate the momentum of the photon emitted in the second transition. kg-m/ kg-m/t kg-m/ kg-m/s kg-m/Explanation / Answer
for hydrogen atom..
energy of electron in nth state is En = -13.6eV/n^2..
in 5th state..E5 = -13.6/25 = -0.544eV..
in 2nd state E2 = -13.6/4 = -3.4eV..
for n = 1 state E1 = -13.6 eV..
for the tranisition from n=5 to n=2...
E5-E2 = 0.544+3.4 = 2.856 eV..
for the transition from n=2 to n=1...
E2-E1 = -3.4+13.6 = 10.2eV...
10.2eV >2.856eV.........(1)..
wavelength of the tranisition is lamda from n=5 to n=2= E5-E2 = h*c/lamda..
lamda = (h*c)/(E5-E2) = 6.625*10^-34*3*10^8/(2.856*1.6*10^-19) = 434.93nm...
wavelength of the tranisition is lamda from n=2 to n=1 = E2-E1 = h*c/lamda..
lamda = (h*c)/(E5-E2) = 6.625*10^-34*3*10^8/(10.2*1.6*10^-19) = 121.78nm...
434.93nm>121.78nm...(2)..
from euations (1) and(2)..
we can say the answer C) in the first transition energy og=f photon is less and the wavelength of the photon is greater..
2) Frequency f= (E5-E2)/h = 2.856*1.6*10^-19/6.625*10^-34 = 6.9*10^14 Hz...
3) momentum p = h/lamda = 6.625*10^-34/121.78 = 5.44*10^-27 kg.m/sec
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