To find the kinetic energy of the emitted photoelectrons and the stopping potential when light of wavelength [tex]\(5 \times 10^{-7}\)[/tex] meters falls on a metal plate with a work function of 1.90 eV, we will follow these steps:
### Step 1: Calculate the Energy of the Incident Photons
The energy of a photon ([tex]\(E\)[/tex]) can be calculated using the formula:
[tex]\[ E = \frac{h \cdot c}{\lambda} \][/tex]
where
- [tex]\( h = 6.62 \times 10^{-34} \)[/tex] Js (Planck's constant),
- [tex]\( c = 3 \times 10^8 \)[/tex] m/s (speed of light),
- [tex]\( \lambda = 5 \times 10^{-7} \)[/tex] m (wavelength).
Plugging in the values:
[tex]\[ E = \frac{6.62 \times 10^{-34} \, \text{Js} \times 3 \times 10^8 \, \text{m/s}}{5 \times 10^{-7} \, \text{m}} \][/tex]
[tex]\[ E = 3.972 \times 10^{-19} \, \text{J} \][/tex]
### Step 2: Convert the Energy from Joules to Electron Volts
To convert the energy from Joules to electron volts (eV), use:
[tex]\[ 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \][/tex]
So,
[tex]\[ \text{Energy in eV} = \frac{3.972 \times 10^{-19} \, \text{J}}{1.6 \times 10^{-19} \, \text{J/eV}} \][/tex]
[tex]\[ \text{Energy in eV} \approx 2.4825 \, \text{eV} \][/tex]
### Step 3: Calculate the Kinetic Energy of the Emitted Photoelectrons
The kinetic energy ([tex]\(E_k\)[/tex]) of the emitted photoelectrons can be calculated using the photoelectric equation:
[tex]\[ E_k = E_{\text{photon}} - \phi \][/tex]
where [tex]\(\phi\)[/tex] is the work function of the metal in eV.
[tex]\[ \phi = 1.90 \, \text{eV} \][/tex]
So,
[tex]\[ E_k = 2.4825 \, \text{eV} - 1.90 \, \text{eV} \][/tex]
[tex]\[ E_k = 0.5825 \, \text{eV} \][/tex]
Therefore, the kinetic energy of the emitted photoelectrons is [tex]\(0.5825\)[/tex] eV.
### Step 4: Calculate the Stopping Potential
The stopping potential ([tex]\(V_s\)[/tex]) is equal to the kinetic energy of the emitted photoelectrons in electron volts:
[tex]\[ V_s = E_k \][/tex]
[tex]\[ V_s = 0.5825 \, \text{V} \][/tex]
### Summary
- The energy of the incident photons is approximately [tex]\(3.972 \times 10^{-19}\)[/tex] Joules.
- The energy of the incident photons in electron volts is approximately [tex]\(2.4825\)[/tex] eV.
- The kinetic energy of the emitted photoelectrons is [tex]\(0.5825\)[/tex] eV.
- The stopping potential is [tex]\(0.5825\)[/tex] V.
Hence, the answer is:
[tex]\[ E_k = 0.5825 \, \text{eV}, \quad V_s = 0.5825 \, \text{V} \][/tex]
