Question

A photon With a wavelength of 397nm is emitted from an electron in an energy level 7 of hydrogen atom. What is the new energy level of the electron?

Answer 1

To determine the new energy level of the electron in the hydrogen atom after emitting a photon with a wavelength of 397 nm, we will use the following steps:

1. **Calculate the energy of the photon** using the formula:

\[
E = \frac{hc}{\lambda}
\]

where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant \((6.626 \times 10^{-34} \, \text{J s})\),
- \(c\) is the speed of light \((3.00 \times 10^8 \, \text{m/s})\),
- \(\lambda\) is the wavelength of the photon in meters (397 nm = \(397 \times 10^{-9} \, \text{m}\)).

2. **Plug in the values**:

\[
E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3.00 \times 10^8 \, \text{m/s})}{397 \times 10^{-9} \, \text{m}}
\]

\[
E \approx \frac{1.98678 \times 10^{-25} \, \text{J m}}{397 \times 10^{-9} \, \text{m}} \approx 5.0005 \times 10^{-19} \, \text{J}
\]

3. **Convert this energy to electron volts (eV)** using the conversion \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\):

\[
E \approx \frac{5.0005 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \, \text{J/eV}} \approx 3.12 \, \text{eV}
\]

4. **Determine the initial energy level of the electron**. The energy levels of a hydrogen atom are given by the formula:

\[
E_n = -\frac{13.6 \, \text{eV}}{n^2}
\]

For an initial energy level \(n = 7\):

\[
E_7 = -\frac{13.6}{7^2} = -\frac{13.6}{49} \approx -0.2776 \, \text{eV}
\]

5. **Calculate the new energy level after the photon is emitted**:

The energy of the final state must be equal to the initial energy minus the energy of the photon. Thus:

\[
E_{\text{final}} = E_7 - E_{photon}
\]

\[
E_{\text{final}} = -0.2776 \, \text{eV} - 3.12 \, \text{eV} \approx -3.3976 \, \text{eV}
\]

6. **Now, determine the new energy level \(n\)** where the energy \(E_n = -3.3976 \, \text{eV}\) corresponds to:

\[
-\frac{13.6}{n^2} = -3.3976
\]

This leads to:

\[
\frac{13.6}{3.3976} \approx n^2
\]

Calculating \(n^2\):

\[
n^2 \approx \frac{13.6}{3.3976} \approx 4
\]

Therefore,

\[
n \approx 2
\]

Thus, the new energy level of the electron after emitting the photon of wavelength 397 nm is **2**.