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The work function of a photoelectric material is 4 eV. (a) What is the threshold wavelength? (b) Find the wavelength of light for which the stopping potential is $2.5{\text{ }}V.$

Last updated date: 19th Jun 2024
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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

${\lambda _{thesh}} = \dfrac{{hc}}{\phi }$
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 ${\lambda _{thesh}} = 190\,nm$