Answered: Find the kinetic, potential, and total…

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Question

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.

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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