A hydrogen atom is in its second excitedstate ( n = 3). Using the Bohrtheory of

Question

A hydrogen atom is in its second excitedstate (n = 3). Using the Bohrtheory of the atom, calculate the following. (a) the radius of the orbit
1 nm

(b) the linear momentum of the electron
2 kg·m/s

(c) the angular momentum of the electron
3 J·s

(d) the kinetic energy
4 eV

(e) the potential energy
5 eV

(f) the total energy
6 eV (a) the radius of the orbit
1 nm

(b) the linear momentum of the electron
2 kg·m/s

(c) the angular momentum of the electron
3 J·s

(d) the kinetic energy
4 eV

(e) the potential energy
5 eV

(f) the total energy
6 eV

Explanation / Answer

(a) the radius of the orbit r = 0.529 * n^ 2
                                        = 4.761 Ao                                         = 0.4761 nm
(b) the linear momentum of the electron P = mv where v = velocity = (2.118 * 10 ^6 m / s) / n              = 706 * 10 ^ 3 m / s            m = mass = 9.11 * 10 ^ -31 kg So P = 6.43* 10 ^ -25 kg m / s

(c) the angular momentum of the electron L = rP                                                                   = 3.062 * 10 ^ -34 kg m^ 2 / s (d) the kinetic energy  = 2 * 2.179* 10 ^ -11 / n^2 erg                            = 2 * 0.242 * 10 ^ -11 erg                            = 2 * 0.242 * 10 ^ -18 J    since 1 erg = 10 ^-7 J                             = 2 *(0.242 * 10 ^ -18 ) / ( 1.6 * 10 ^ -19 ) eV                             =  3.026eV

(e) the potential energy =2.179* 10 ^ -11 / n^ 2 erg                            = 0.242 * 10 ^ -11 erg                            = 0.242 * 10 ^ -18 J    since 1 erg = 10 ^ -7J                             = (0.242 * 10 ^ -18 ) / ( 1.6 * 10 ^ -19 ) eV                             =  1.513 eV

(f) the total energy = -2.179* 10 ^ -11 / n^ 2 erg                            = 2 * 0.242 * 10 ^ -11 erg                            = 2 * 0.242 * 10 ^ -18 J    since 1 erg = 10 ^-7 J                             = 2 *(0.242 * 10 ^ -18 ) / ( 1.6 * 10 ^ -19 ) eV                             =  3.026eV
                           = 0.242 * 10 ^ -11 erg                            = 0.242 * 10 ^ -18 J    since 1 erg = 10 ^ -7J                             = (0.242 * 10 ^ -18 ) / ( 1.6 * 10 ^ -19 ) eV                             =  1.513 eV
                           = -0.242 * 10 ^ -11 erg                            = -0.242 * 10 ^ -18 J    since 1 erg = 10 ^ -7J                             = (-0.242 * 10 ^ -18 ) / ( 1.6 * 10 ^ -19 ) eV                             = -1.513 eV

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