The work function of metal is in the range of \[{\text{2eV}}\]to \[{\text{5eV}}\]. Find which of the following wavelengths of light cannot be used for photoelectric effect. (Consider, Planck constant = \[{\text{4}} \times {\text{1}}{{\text{0}}^{ - 15}}{\text{eVs}}\], velocity of light = \[3 \times {\text{1}}{{\text{0}}^8}m/s\]).
A. \[510nm\]
B. \[650nm\]
C. \[400nm\]
D. \[570nm\]
Answer
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Hint: Planck's constant is the physical constant in the quantum of electromagnetism that relates the energy carried by a single photon to its corresponding frequency. Described by h and weigh, using J.s in SI and eV.s in MKS. To quantum mechanics, the value of Planck 's constant has prime significance.
Complete step-by-step answer:
Minimum wavelength resulting in photoelectric effect = \[\dfrac{{hc}}{{{E_{max}}}}\]
= \[\dfrac{{4 \times {{10}^{ - 15}} \times 3 \times {{10}^8}}}{5}m\]
= \[240nm\]
Maximum wavelength resulting in photoelectric effect = \[\dfrac{{hc}}{{{E_{min}}}}\]
= \[\dfrac{{4 \times {{10}^{ - 15}} \times 3 \times {{10}^8}}}{2}m\]
= \[600nm\]
Therefore, wavelength light \[650nm\] is not acceptable.
Note: The photoelectric effect is a phenomenon in which electrons are expelled from a metal's surface when light occurs thereon. Those electrons being ejected are called photoelectrons. It is important to note that the photoelectron emission and the kinetic energy of the ejected photoelectrons depend on the light frequency that occurs on the surface of the metal. Photoemission is generally referred to as the mechanism by which photoelectrons are expelled from the metal surface due to the action of light. One cannot clarify the photoelectric effect by considering light as a wave. This phenomenon can also be explained by the particle essence of light, in which light can be visualized as a stream of electromagnetic energy particles. These luminous 'particles' are called photons. The energy a photon holds is related to the light frequency through Planck 's equation. So, it can be understood that photons with similar energies bear different wavelengths of light. For example, the frequency of blue light is greater than that of red light (the blue light 's wavelength is much shorter than the red light's wavelength). Therefore, the energy held by a blue light photon would be greater than the energy held by a red-light photon.
Complete step-by-step answer:
Minimum wavelength resulting in photoelectric effect = \[\dfrac{{hc}}{{{E_{max}}}}\]
= \[\dfrac{{4 \times {{10}^{ - 15}} \times 3 \times {{10}^8}}}{5}m\]
= \[240nm\]
Maximum wavelength resulting in photoelectric effect = \[\dfrac{{hc}}{{{E_{min}}}}\]
= \[\dfrac{{4 \times {{10}^{ - 15}} \times 3 \times {{10}^8}}}{2}m\]
= \[600nm\]
Therefore, wavelength light \[650nm\] is not acceptable.
Note: The photoelectric effect is a phenomenon in which electrons are expelled from a metal's surface when light occurs thereon. Those electrons being ejected are called photoelectrons. It is important to note that the photoelectron emission and the kinetic energy of the ejected photoelectrons depend on the light frequency that occurs on the surface of the metal. Photoemission is generally referred to as the mechanism by which photoelectrons are expelled from the metal surface due to the action of light. One cannot clarify the photoelectric effect by considering light as a wave. This phenomenon can also be explained by the particle essence of light, in which light can be visualized as a stream of electromagnetic energy particles. These luminous 'particles' are called photons. The energy a photon holds is related to the light frequency through Planck 's equation. So, it can be understood that photons with similar energies bear different wavelengths of light. For example, the frequency of blue light is greater than that of red light (the blue light 's wavelength is much shorter than the red light's wavelength). Therefore, the energy held by a blue light photon would be greater than the energy held by a red-light photon.
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