Let's solve the problem step-by-step.
### Step 1: Convert the wavelength from Angstroms to meters
Given wavelength [tex]\( \lambda = 3000 \, \text{Å} \)[/tex].
We know that [tex]\( 1 \, \text{Å} = 10^{-10} \, \text{meters} \)[/tex].
So,
[tex]\[
\lambda = 3000 \times 10^{-10} = 3 \times 10^{-7} \, \text{meters}
\][/tex]
### Step 2: Convert the work function from eV to Joules
Given work function [tex]\( \phi = 2.20 \, \text{eV} \)[/tex].
We know that [tex]\( 1 \, \text{eV} = 1.60218 \times 10^{-19} \, \text{Joules} \)[/tex].
So,
[tex]\[
\phi = 2.20 \times 1.60218 \times 10^{-19} = 3.524796 \times 10^{-19} \, \text{Joules}
\][/tex]
### Step 3: Calculate the energy of the photon
Energy of a photon is given by the equation [tex]\( E = \frac{hc}{\lambda} \)[/tex],
where:
- [tex]\( h \)[/tex] is Planck's constant ([tex]\( h = 6.62607015 \times 10^{-34} \, \text{J} \cdot \text{s} \)[/tex]),
- [tex]\( c \)[/tex] is the speed of light ([tex]\( c = 3.00 \times 10^8 \, \text{m/s} \)[/tex]),
- [tex]\( \lambda \)[/tex] is the wavelength ([tex]\( 3 \times 10^{-7} \, \text{meters} \)[/tex]).
So,
[tex]\[
E = \frac{6.62607015 \times 10^{-34} \times 3.00 \times 10^8}{3 \times 10^{-7}} = 6.62607015 \times 10^{-19} \, \text{Joules}
\][/tex]
### Step 4: Calculate the kinetic energy of the emitted photoelectron
Using the photoelectric equation:
[tex]\[
\text{Kinetic Energy} = \text{Energy of Photon} - \text{Work Function}
\][/tex]
[tex]\[
\text{Kinetic Energy} = 6.62607015 \times 10^{-19} - 3.524796 \times 10^{-19} = 3.10127415 \times 10^{-19} \, \text{Joules}
\][/tex]
So, the kinetic energy of the emitted photoelectron is [tex]\( 3.10127415 \times 10^{-19} \, \text{Joules} \)[/tex]. This value is closest to the provided option:
[tex]\[
\boxed{3.08 \times 10^{-19} \, \text{J}}
\][/tex]
