Question

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.

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Answer 1

(a) In the ground state the energy of the electron is minimum. Also energy E $\propto$ -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 $\pi$ ^2mZe^2

r (n = 1) = 1 * h^2/4 $\pi$ ^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