Our understanding of the H atom will help us learn about atoms with more electrons. The n =1 electron energy level of a H atom has an energy of 2.18  10–18 J. (a) What is the energy of the n = 5 level? (b) Calculate the wavelength and frequency of a photon emitted when an electron jumps down from n = 5 to n = 1 in a H atom.

Respuesta :

Answer:

a) What is the energy of the n = 5 level?

[tex]E_{5} = - 8.70x10^{-20}J[/tex]

(b) Calculate the wavelength and frequency of a photon emitted when an electron jumps down from n = 5 to n = 1 in a H atom.

[tex]\lambda = 94.9nm[/tex], [tex]f = 3.16x10^{15}Hz[/tex]

Explanation:

The permitted energy for the atom of hydrogen according with the Bohr's model is defined as:

[tex]E_{n} = -\frac{13.606 eV}{n^{2}}[/tex]  (1)

Or it can be expressed in Joules, since [tex]1eV = 1.60x10^{-19}J[/tex]

[tex]E_{n} = -\frac{2.18x10^{-18} J}{n^{2}}[/tex]

Where the value [tex]-2.18x10^{-18}[/tex] represents the energy of the ground state¹ and n is the principal quantum number.

a) What is the energy of the n = 5 level?

For the case of [tex]n = 5[/tex]:

[tex]E_{5} = -\frac{2.18x10^{-18} J}{(5)^{2}}[/tex]

[tex]E_{5} = -8.70x10^{-20}J[/tex]

So the energy of the [tex]n = 5[/tex] level is [tex]-8.70x10^{-20}J[/tex].   

(b) Calculate the wavelength and frequency of a photon emitted when an electron jumps down from n = 5 to n = 1 in a H atom.

The wavelength can be determined by means of the Rydberg formula:

[tex]\frac{1}{\lambda} = R(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})[/tex]  (2)

Where R is the Rydberg constant, with a value of [tex]1.097x10^{7}m^{-1}[/tex]

For this particular case [tex]n_{f} = 1[/tex] and [tex]n_{i} = 5[/tex]:

[tex]\frac{1}{\lambda} = 1.097x10^{7}m^{-1}(\frac{1}{(1)^{2}}-\frac{1}{(5)^{2}})[/tex]

[tex]\frac{1}{\lambda} = 1.097x10^{7}m^{-1}(0.96)[/tex]

[tex]\frac{1}{\lambda} = 10531200m^{-1}[/tex]

[tex]\lambda = \frac{1}{10531200m^{-1}}[/tex]

[tex]\lambda = 9.49x10^{-8}m[/tex]

Where [tex]1m = 1x10^{9}nm[/tex]

[tex]\lambda = 9.49x10^{-8}m[/tex] .  [tex]\frac{1x10^{9}nm}{1m}[/tex]

[tex]\lambda = 94.9nm[/tex]

The frequency can be found by means of:

[tex]c = f\lambda[/tex]   (3)

Equation (3) can be rewritten in terms of [tex]f[/tex]:

[tex]f = \frac{c}{\lambda}[/tex]

[tex]f = \frac{3.00x10^{8}m/s}{9.49x10^{-8}m}[/tex]

[tex]f = 3.16x10^{15}s^{-1}[/tex]

Where [tex]1Hz = s^{-1}[/tex]

[tex]f = 3.16x10^{15}Hz[/tex]

Key terms:

¹Ground state: State of minimum energy.  

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