The energy required to remove an electron from sodium 2.3 eV. Does sodium show a photoelectric effect from orange light with lemda = 6800 Å.
ANSWER : \(\text{Ionization Energy of Sodium:}\)
The ionization energy of sodium is approximately \(2.3\) electron volts (eV).
\(\text{Calculating Energy of Orange Light:}\)
The energy of light can be determined using the equation:
\[E = \frac{hc}{\lambda}\]
Where:
\(E\) is the energy of the light,
\(h = 6.626 \times 10^{-34}\) Js (Planck's constant),
\(c = 3.00 \times 10^8\) m/s (speed of light), and
\(\lambda = 6800 \times 10^{-10}\) m (wavelength of light).
Given that the wavelength of orange light is \(6800\) angstroms, we convert it to meters (\(6800 \times 10^{-10}\) m)). Substituting these values into the equation:
\[E = \frac{(6.626 \times 10^{-34} \text{ Js}) \times (3.00 \times 10^8 \text{ m/s})}{6800 \times 10^{-10} \text{ m}}\]
\[E \approx 2.91 \text{ eV}\]
\(\text{Interpretation:}\)
This calculation shows that orange light with a wavelength of \(6800\) angstroms has an energy of approximately \(2.91\) eV. Because this energy is higher than the ionization energy of sodium (\(2.3\) eV), sodium can experience the photoelectric effect when exposed to orange light.
\(\text{Photoelectric Effect:}\)
The photoelectric effect occurs when light, in this case, orange light, interacts with a material, like sodium, and ejects electrons from its surface due to the energy carried by the photons in the light.
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