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To solve the problem of finding the stopping potential when light is used to produce photoelectric emission from the surface of a metal with a given work function, we can follow these steps:
### Step 1: Determine the wavelength of the light wave
From the given electric field equation for the light wave:
[tex]\[ E = E_0 \sin \left( 1.57 \times 10^7 (x - ct) \right) \][/tex]
The term [tex]\( 1.57 \times 10^7 \)[/tex] in the argument of the sine function is the angular wave number [tex]\( k \)[/tex] and is given in radians per meter.
[tex]\[ k = 1.57 \times 10^7 \, \text{rad/m} \][/tex]
The relationship between wave number [tex]\( k \)[/tex] and wavelength [tex]\( \lambda \)[/tex] is:
[tex]\[ k = \frac{2\pi}{\lambda} \][/tex]
Rearranging this for [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = \frac{2\pi}{k} \][/tex]
Plugging in the value of [tex]\( k \)[/tex]:
[tex]\[ \lambda = \frac{2\pi}{1.57 \times 10^7} \approx 4.002 \times 10^{-7} \, \text{m} \][/tex]
### Step 2: Calculate the energy of the photon
The energy [tex]\( E_{\text{photon}} \)[/tex] of a photon is related to its wavelength [tex]\( \lambda \)[/tex] by the equation:
[tex]\[ E_{\text{photon}} = \frac{hc}{\lambda} \][/tex]
Where:
- [tex]\( h \)[/tex] is Planck’s constant ([tex]\( 6.62607015 \times 10^{-34} \, \text{Js} \)[/tex])
- [tex]\( c \)[/tex] is the speed of light ([tex]\( 3.0 \times 10^8 \, \text{m/s} \)[/tex])
Substituting the known values:
[tex]\[ E_{\text{photon}} = \frac{6.62607015 \times 10^{-34} \times 3.0 \times 10^8}{4.002 \times 10^{-7}} \approx 4.967 \times 10^{-19} \, \text{J} \][/tex]
### Step 3: Convert the work function from eV to Joules
The work function [tex]\( \phi \)[/tex] is given as [tex]\( 1.9 \, \text{eV} \)[/tex]. To convert electron volts to Joules:
[tex]\[ 1 \, \text{eV} = 1.60218 \times 10^{-19} \, \text{J} \][/tex]
Thus:
[tex]\[ \phi = 1.9 \, \text{eV} \times 1.60218 \times 10^{-19} \, \text{J/eV} \approx 3.044 \times 10^{-19} \, \text{J} \][/tex]
### Step 4: Calculate the maximum kinetic energy of the emitted electron
Using the photoelectric effect equation, the kinetic energy [tex]\( KE_{\text{max}} \)[/tex] of the emitted electron is given by:
[tex]\[ KE_{\text{max}} = E_{\text{photon}} - \phi \][/tex]
Substituting in the values:
[tex]\[ KE_{\text{max}} = 4.967 \times 10^{-19} \, \text{J} - 3.044 \times 10^{-19} \, \text{J} \approx 1.923 \times 10^{-19} \, \text{J} \][/tex]
### Step 5: Determine the stopping potential
The stopping potential [tex]\( V_{\text{stop}} \)[/tex] is related to the maximum kinetic energy by:
[tex]\[ KE_{\text{max}} = eV_{\text{stop}} \][/tex]
Where [tex]\( e \)[/tex] is the elementary charge ([tex]\( 1.60218 \times 10^{-19} \, \text{C} \)[/tex]).
Solving for [tex]\( V_{\text{stop}} \)[/tex]:
[tex]\[ V_{\text{stop}} = \frac{KE_{\text{max}}}{e} \][/tex]
[tex]\[ V_{\text{stop}} = \frac{1.923 \times 10^{-19} \, \text{J}}{1.60218 \times 10^{-19} \, \text{C}} \approx 1.200 \, \text{V} \][/tex]
### Conclusion
The stopping potential for the photoelectric emission from the metal surface is approximately 1.20 V.
### Step 1: Determine the wavelength of the light wave
From the given electric field equation for the light wave:
[tex]\[ E = E_0 \sin \left( 1.57 \times 10^7 (x - ct) \right) \][/tex]
The term [tex]\( 1.57 \times 10^7 \)[/tex] in the argument of the sine function is the angular wave number [tex]\( k \)[/tex] and is given in radians per meter.
[tex]\[ k = 1.57 \times 10^7 \, \text{rad/m} \][/tex]
The relationship between wave number [tex]\( k \)[/tex] and wavelength [tex]\( \lambda \)[/tex] is:
[tex]\[ k = \frac{2\pi}{\lambda} \][/tex]
Rearranging this for [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = \frac{2\pi}{k} \][/tex]
Plugging in the value of [tex]\( k \)[/tex]:
[tex]\[ \lambda = \frac{2\pi}{1.57 \times 10^7} \approx 4.002 \times 10^{-7} \, \text{m} \][/tex]
### Step 2: Calculate the energy of the photon
The energy [tex]\( E_{\text{photon}} \)[/tex] of a photon is related to its wavelength [tex]\( \lambda \)[/tex] by the equation:
[tex]\[ E_{\text{photon}} = \frac{hc}{\lambda} \][/tex]
Where:
- [tex]\( h \)[/tex] is Planck’s constant ([tex]\( 6.62607015 \times 10^{-34} \, \text{Js} \)[/tex])
- [tex]\( c \)[/tex] is the speed of light ([tex]\( 3.0 \times 10^8 \, \text{m/s} \)[/tex])
Substituting the known values:
[tex]\[ E_{\text{photon}} = \frac{6.62607015 \times 10^{-34} \times 3.0 \times 10^8}{4.002 \times 10^{-7}} \approx 4.967 \times 10^{-19} \, \text{J} \][/tex]
### Step 3: Convert the work function from eV to Joules
The work function [tex]\( \phi \)[/tex] is given as [tex]\( 1.9 \, \text{eV} \)[/tex]. To convert electron volts to Joules:
[tex]\[ 1 \, \text{eV} = 1.60218 \times 10^{-19} \, \text{J} \][/tex]
Thus:
[tex]\[ \phi = 1.9 \, \text{eV} \times 1.60218 \times 10^{-19} \, \text{J/eV} \approx 3.044 \times 10^{-19} \, \text{J} \][/tex]
### Step 4: Calculate the maximum kinetic energy of the emitted electron
Using the photoelectric effect equation, the kinetic energy [tex]\( KE_{\text{max}} \)[/tex] of the emitted electron is given by:
[tex]\[ KE_{\text{max}} = E_{\text{photon}} - \phi \][/tex]
Substituting in the values:
[tex]\[ KE_{\text{max}} = 4.967 \times 10^{-19} \, \text{J} - 3.044 \times 10^{-19} \, \text{J} \approx 1.923 \times 10^{-19} \, \text{J} \][/tex]
### Step 5: Determine the stopping potential
The stopping potential [tex]\( V_{\text{stop}} \)[/tex] is related to the maximum kinetic energy by:
[tex]\[ KE_{\text{max}} = eV_{\text{stop}} \][/tex]
Where [tex]\( e \)[/tex] is the elementary charge ([tex]\( 1.60218 \times 10^{-19} \, \text{C} \)[/tex]).
Solving for [tex]\( V_{\text{stop}} \)[/tex]:
[tex]\[ V_{\text{stop}} = \frac{KE_{\text{max}}}{e} \][/tex]
[tex]\[ V_{\text{stop}} = \frac{1.923 \times 10^{-19} \, \text{J}}{1.60218 \times 10^{-19} \, \text{C}} \approx 1.200 \, \text{V} \][/tex]
### Conclusion
The stopping potential for the photoelectric emission from the metal surface is approximately 1.20 V.
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