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

The energy of an electron in the hydrogen atom is quantized, and the energy of the n = 5 level can be calculated using the equation En = (-13.6 eV/n^2). When an electron transitions from the n = 5 to the n=1 energy level, a photon is emitted.

Step-by-step explanation:

The energy of an electron in the hydrogen atom is quantized and can only have discrete values given by the equation En = (-13.6 eV/n^2). To find the energy of the n = 5 level, we can substitute n = 5 into the equation and calculate the energy.

For Part B, when an electron transitions from the n = 5 to the n = 1 energy level, a photon is emitted. The energy of this photon can be calculated using the equation AE = hf = E1 - Ef, where E1 is the initial energy and Ef is the final energy.

Answer 2

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`

Step-by-step explanation:

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

`E_(n) = -(13.606 eV)/(n^(2))` (1)

Or it can be expressed in Joules, since
`1eV = 1.60x10^(-19)J`

`E_(n) = -(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) = -(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:

`(1)/(\lambda) = R((1)/(n_(f)^(2))-(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`:

`(1)/(\lambda) = 1.097x10^(7)m^(-1)((1)/((1)^(2))-(1)/((5)^(2)))`

`(1)/(\lambda) = 1.097x10^(7)m^(-1)(0.96)`

`(1)/(\lambda) = 10531200m^(-1)`

`\lambda = (1)/(10531200m^(-1))`

`\lambda = 9.49x10^(-8)m`

Where
`1m = 1x10^(9)nm`

`\lambda = 9.49x10^(-8)m` .
`(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 = (c)/(\lambda)`

`f = (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.