
The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal (for which work function is 1.65 \[eV\]) will be ________.
A. \[4 \times {10^{10}}Hz\]
B. \[4 \times {10^{11}}Hz\]
C. \[4 \times {10^{14}}Hz\]
D. \[4 \times {10^{ - 10}}Hz\]
Answer
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Hint: Threshold frequency is the minimum frequency required to eject electrons from a metal surface. It is the characteristics of a metal. Work function is defined as the minimum energy required to move an electron from to infinity from the surface of the metal.
Formula used Work function is given as:
\[\phi = h{\upsilon _0} = \dfrac{{hc}}{{{\lambda _0}}}\]
Where h is Planck’s constant, c is the speed of light, \[{\upsilon _0}\] is the frequency of incident light and \[{\lambda _0}\] is the threshold frequency.
1eV=\[1.6 \times {10^{ - 19}}J\]
Complete step by step solution:
Given Work function, \[\phi = \] 1.65 \[eV\]
As we know Work function is given as,
\[\phi = h{\upsilon _0}\]
\[\Rightarrow {\upsilon _0} = \dfrac{\phi }{h}\]
\[\Rightarrow {\upsilon _0} = \dfrac{{1.65 \times 1.6 \times {{10}^{ - 19}}}}{{6.63 \times {{10}^{ - 34}}}}\]
\[\therefore {\upsilon _0} \approx 4 \times {10^{14}}Hz\]
Therefore, the lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal (for which work function is 1.65 \[eV\]) will be \[4 \times {10^{14}}Hz\].
Hence option C is the correct answer.
Additional information: The photoelectric effect is the phenomenon in which when light arises on a metal surface then the electrons are ejected from it. These electrons released are known as photoelectrons. Photon energy is the energy carried by a single photon. The energy is directly related to the photon's electromagnetic frequency and is inversely related to the wavelength. The higher the photon's frequency the energy will be higher.
Note: Most of the students have confusion between the incident frequency and threshold frequency. Always remember that the threshold frequency is the characteristic of a metal and differs for different metals while the incident frequency depends on the energy of the incident radiation.
Formula used Work function is given as:
\[\phi = h{\upsilon _0} = \dfrac{{hc}}{{{\lambda _0}}}\]
Where h is Planck’s constant, c is the speed of light, \[{\upsilon _0}\] is the frequency of incident light and \[{\lambda _0}\] is the threshold frequency.
1eV=\[1.6 \times {10^{ - 19}}J\]
Complete step by step solution:
Given Work function, \[\phi = \] 1.65 \[eV\]
As we know Work function is given as,
\[\phi = h{\upsilon _0}\]
\[\Rightarrow {\upsilon _0} = \dfrac{\phi }{h}\]
\[\Rightarrow {\upsilon _0} = \dfrac{{1.65 \times 1.6 \times {{10}^{ - 19}}}}{{6.63 \times {{10}^{ - 34}}}}\]
\[\therefore {\upsilon _0} \approx 4 \times {10^{14}}Hz\]
Therefore, the lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal (for which work function is 1.65 \[eV\]) will be \[4 \times {10^{14}}Hz\].
Hence option C is the correct answer.
Additional information: The photoelectric effect is the phenomenon in which when light arises on a metal surface then the electrons are ejected from it. These electrons released are known as photoelectrons. Photon energy is the energy carried by a single photon. The energy is directly related to the photon's electromagnetic frequency and is inversely related to the wavelength. The higher the photon's frequency the energy will be higher.
Note: Most of the students have confusion between the incident frequency and threshold frequency. Always remember that the threshold frequency is the characteristic of a metal and differs for different metals while the incident frequency depends on the energy of the incident radiation.
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