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Hint: The threshold wavelength corresponds to the work function of a photoelectric material and is inversely proportional to it. The stopping corresponds to the potential required to stop electrons ejected from the photoelectric material.
Formula used: In this solution, we will use the following formula:
$\phi = \dfrac{{hc}}{{{\lambda _{thresh}}}}$ where $h$ is the Planck’s constant, $c$ is the speed of light, and ${\lambda _{thresh}}$ is the wavelength of the photon
Complete step by step answer:
We’ve been given the work function of a photoelectric material as 4 eV. The threshold wavelength of the material corresponds to the energy that the incoming photon must have to cause the photoelectric effect. Since the incoming photon must have energy equal to the work function of the material, we can calculate the threshold wavelength as
\[{\lambda _{thesh}} = \dfrac{{hc}}{\phi }\]
Substituting the value of $h = 6.63 \times {10^{ - 34}}$, $c = 3 \times {10^8}$ and $\phi = 4\, \times 1.6 \times {10^{ - 19}}V$$(\because e = 1.6 \times {10^{ - 19}})$, we get
\[{\lambda _{thesh}} = 3.1 \times {10^{ - 7}}\,m\] or equivalently \[{\lambda _{thesh}} = 310\,nm\]
Hence the threshold wavelength of the material is 310 nm. The incoming photon must have a wavelength less than or equal to this value.
b) Now we know that the stopping potential is \[2.5{\text{ }}V.\] and we want to find the corresponding threshold wavelength. So again, using the formula
\[{\lambda _{thesh}} = \dfrac{{hc}}{\phi }\]
Substituting the value of $h = 6.63 \times {10^{ - 34}}$, $c = 3 \times {10^8}$ and $\phi = 2.5 \times 1.6 \times {10^{ - 19}}V$$(\because e = 1.6 \times {10^{ - 19}})$, we get
\[{\lambda _{thesh}} = 1.91 \times {10^{ - 7}}\,m\] or equivalently \[{\lambda _{thesh}} = 190\,nm\]
Note: The threshold wavelength corresponds to the work function of the material and if a photon corresponding to the threshold wavelength is incident on the material, the ejected electron will have no kinetic energy and will eventually recombine with the metal. To have a non-zero ejected electron velocity, the wavelength of the photon must be less than the threshold wavelength.
Formula used: In this solution, we will use the following formula:
$\phi = \dfrac{{hc}}{{{\lambda _{thresh}}}}$ where $h$ is the Planck’s constant, $c$ is the speed of light, and ${\lambda _{thresh}}$ is the wavelength of the photon
Complete step by step answer:
We’ve been given the work function of a photoelectric material as 4 eV. The threshold wavelength of the material corresponds to the energy that the incoming photon must have to cause the photoelectric effect. Since the incoming photon must have energy equal to the work function of the material, we can calculate the threshold wavelength as
\[{\lambda _{thesh}} = \dfrac{{hc}}{\phi }\]
Substituting the value of $h = 6.63 \times {10^{ - 34}}$, $c = 3 \times {10^8}$ and $\phi = 4\, \times 1.6 \times {10^{ - 19}}V$$(\because e = 1.6 \times {10^{ - 19}})$, we get
\[{\lambda _{thesh}} = 3.1 \times {10^{ - 7}}\,m\] or equivalently \[{\lambda _{thesh}} = 310\,nm\]
Hence the threshold wavelength of the material is 310 nm. The incoming photon must have a wavelength less than or equal to this value.
b) Now we know that the stopping potential is \[2.5{\text{ }}V.\] and we want to find the corresponding threshold wavelength. So again, using the formula
\[{\lambda _{thesh}} = \dfrac{{hc}}{\phi }\]
Substituting the value of $h = 6.63 \times {10^{ - 34}}$, $c = 3 \times {10^8}$ and $\phi = 2.5 \times 1.6 \times {10^{ - 19}}V$$(\because e = 1.6 \times {10^{ - 19}})$, we get
\[{\lambda _{thesh}} = 1.91 \times {10^{ - 7}}\,m\] or equivalently \[{\lambda _{thesh}} = 190\,nm\]
Note: The threshold wavelength corresponds to the work function of the material and if a photon corresponding to the threshold wavelength is incident on the material, the ejected electron will have no kinetic energy and will eventually recombine with the metal. To have a non-zero ejected electron velocity, the wavelength of the photon must be less than the threshold wavelength.
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