Find the kinetic, potential, and total mechanical energies of the hydrogen atom in the first excited level, and find the wavelength of the photon emitted in a transition from that level to the ground level.

Find the kinetic, potential, and total mechanical energies of the hydrogen atom in the first excited level, and find the wavelength of the photon emitted in a transition from that level to the ground level.

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Question

Expert Solution

Step 1

For a Hydrogen atom, the kinetic energy of a Hydrogen atom is the energy of its revolving electrons, The angular momentum of the electrons can be given as

$m v r = \frac{n h}{2 π} s q u a r i n g b o t h n s i d e s m^{2} v^{2} r^{2} = \frac{n^{2} h^{2}}{4 π^{2}} m^{2} v^{2} = \frac{n^{2} h^{2}}{4 π^{2} r^{2}} d i v i d i n g b o t h t h e s i d e s o f t h e e q u a t i o n b y m m v^{2} = \frac{n^{2} h^{2}}{4 π^{2} {mr}^{2}} t h e k i n e t i c e n e r g y, K = \frac{1}{2} m v^{2} = \frac{n^{2} h^{2}}{8 π^{2} {mr}^{2}} I n t h e 1 s t e x c i t e d s t a t e, n = 2; r_{n} = 0.53 \times \frac{n^{2}}{Z} A r = 0.53 \times 4 = 2.12 A$

Step 2

$K = \frac{4 \times h^{2}}{8 π^{2} {(2.12 \times 10^{- 10})}^{2}} = \frac{0.11 h^{2}}{π^{2} \times 10^{- 20}} = 0.0113 h^{2} \times 10^{20} K = 0.0113 \times {(6.6 \times 10^{- 34})}^{2} = 4.915 \times 10^{- 14} J = 3.072 \times 10^{- 30} e V$

$T h e p o t e n t i a l e n e r g y i s g i v e n b y U (r_{n}) = \frac{1}{4 {πε}_{0}} \frac{e^{2}}{r_{n}} = \frac{9 \times 10^{9} \times {(1.6 \times 10^{- 19})}^{2}}{{(2.12 \times 10^{- 10})}^{2}} = 5.126 \times 10^{- 9} J$

$U = 5.126 \times 10^{- 9} J = 3.20375 \times 10^{10} e V$

$T h e t o t a l M e c h a n i c a l e n e r g y E = K + U = (3.072 \times 10^{- 30} + 3.20375 \times 10^{10}) e V = 3.20375 \times 10^{10} e V$

Answer: Total energy of the system $E = 3.20375 \times 10^{10} e V$

The Total kinetic energy, K= $3.072 \times 10^{- 30} e V$

The total potential energy, U= $3.20375 \times 10^{10} e V$

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