inorganic chemistry help!
Consider the two seperated hydrogen atoms. in the first hydrogen
atom, the electron is in the n=1 Bohr orbit. in the second hydrogen
atom, the electron is in the n=4 Bohr orbit.
picture updated
(a) In the ground state the energy of the electron is minimum.
Also energy E -1/n^2. So,
greater the value of n greater will be the energy. So n=1 is the
ground state.
(b) In the first the electron moves faster.
equate centripetal force with coulomb force between two point
charges.
mv²/r = Ze²/4πεr² and solve for r to get:
r = Ze²/4πεmv²
You are given the equation, mvr = nh/2π, solve for r to get:
r = nh/2πmv
equate both r together to get:
Ze²/4πεmv² = nh/2πmv
solve for v
v = Ze²/2εnh
Z=1 in hydrogen so:
v = (1.6x10^-19)²/[2(8.85x10^-12)(1)(6.63x10^-27)
v = 2.18x10^6 m/s
v (n=4)= Ze²/2ε *4*h
= 2.18x10^6 ms^-1/4 = 0.545 *10^6 m/s
(c) The radius of the n=4 will be larger.
r(n) = n^2h^2/4^2mZe^2
r (n = 1) = 1 * h^2/4^2m * 1 *e^2
= 1 *(6.636 810^-27 erg.s)/ 4 *(3.14 )^2 * (9.11 *10^-28) (4.8*10^-10)^2 = 0.529 A^o
r (n = 4) = n^2 * 0.529 A^o = 8.464 A^o
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