Question

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.

Answer 1

Answer:

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

$E_{5} = - 8.70x10^{-20}J$

(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.

$\lambda = 94.9nm$, $f = 3.16x10^{15}Hz$

Explanation:

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

$E_{n} = -\frac{13.606 eV}{n^{2}}$  (1)

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

$E_{n} = -\frac{2.18x10^{-18} J}{n^{2}}$

Where the value $-2.18x10^{-18}$ 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 $n = 5$:

$E_{5} = -\frac{2.18x10^{-18} J}{(5)^{2}}$

$E_{5} = -8.70x10^{-20}J$

So the energy of the $n = 5$ level is $-8.70x10^{-20}J$.   

(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:

$\frac{1}{\lambda} = R(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}})$  (2)

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

For this particular case $n_{f} = 1$ and $n_{i} = 5$:

$\frac{1}{\lambda} = 1.097x10^{7}m^{-1}(\frac{1}{(1)^{2}}-\frac{1}{(5)^{2}})$

$\frac{1}{\lambda} = 1.097x10^{7}m^{-1}(0.96)$

$\frac{1}{\lambda} = 10531200m^{-1}$

$\lambda = \frac{1}{10531200m^{-1}}$

$\lambda = 9.49x10^{-8}m$

Where $1m = 1x10^{9}nm$

$\lambda = 9.49x10^{-8}m$ .  $\frac{1x10^{9}nm}{1m}$

$\lambda = 94.9nm$

The frequency can be found by means of:

$c = f\lambda$   (3)

Equation (3) can be rewritten in terms of $f$:

$f = \frac{c}{\lambda}$

$f = \frac{3.00x10^{8}m/s}{9.49x10^{-8}m}$

$f = 3.16x10^{15}s^{-1}$

Where $1Hz = s^{-1}$

$f = 3.16x10^{15}Hz$

Key terms:

¹Ground state: State of minimum energy.