3. Consider an electron in the n-2 state of the Bohr model of the hydrogen atom.

Question

3. Consider an electron in the n-2 state of the Bohr model of the hydrogen atom. (a) Using de Broglie's standing wave perspective for the electron in this orbit calculate the magnitude of the magnetic field produced at the nucleus due to this electron's circulation. (b) What is the wavelength and linear momentum of the photon emitted if this electron makes a transition to the ground state? (c) Calculate the recoil speed of the hydrogen atom due to this transition. (d) An atom in the p-subshell of the L-shell, emits a photon of wavelength 600-nm as it makes a transition to the s-subshell. What is the principal energy level of this atom? (e) What is the angular momentum of the atom? () An electron in an atom drops from the L shell to the K shell and gives an X-ray with wavelength 0.0279 nm. What is atomic number of this atom?

Explanation / Answer

for the hydrogen atom in n = 2 state

a. from debroglie's perespective of standing waves

n*lambda = 2*pi*r

also

ke^2/r = mv^2

also

mvr = nh/2*pi

hence

mv*n*lambda/2*pi = nh/2*pi

mv*lambda = h

mv = h/lambda

mvke^2/mv^2 = nh/2*pi

2*pi*ke^2/nh = v

hence magnetic field at nucleus = muo*q/T*r

T = 2*pi*r/v

B = muo*e*v/2*pi*r^2

B = mu*e*pi*ke^2*mv*pi*mv/nh*(nh)(nh)

a. B = mu*pi^2*k*m^2*e^3/n^3*h^3

B = 4*pi*10^(-7)*pi^2*8.98*10^9*(9.1*10^-31)^2*(1.6*10^-19)^3/27*(6.62*10^(-34))^3

B = 4.822*10^-14 T

b. En = -(m/2)(4*pi^2*k^2*e^4/n^2h^2)

En = -(2m*pi^2*k^2*e^4/n^2h^2)

hence

E(1) - E(3) = 2*m*pi^2*k^2*e^4/h^2 (1/1 - 1/9) = 1.91965*10^-18 J

hecne dE = hc/lambda

lambda = 1.034*10^-7 m

linear momentum = mdv = m(2*pi*ke^2/h)(1 - 1/9) = 1.76492*10^-24 kg m/s

c. recoil speed of hydrogen atom = 1.76492*10^-24 / 1.6*10^-27 = 1103.08014648 m/s

d. the principal energy level of the atom is the energy level of the L shell

n = 2 for L shell, hence

E(2) = 2*m*pi^2*k^2*e^4/h^2 *4 = 5.399015625*10^-19 J

e. angular momentum L changes from 1 to 0

hence angular momentum of the atom is given by L = (h/2*pi)sqrt(l(l + 1)) = (h/2*pi)sqrt(2) = 1.4922*10^-34 kg m^2/s

f. Z^2*2*m*pi^2*k^2*e^4/h^2 (1/1 - 1/4) = hc/0.02779*10^-9

Z = 66.47

hence atomic number is apx 66

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