If work function of metal is 3 eV then threshold wavelength will be
A.\[4125\,\mathop A\limits^0 \]
B. \[4000\,\mathop A\limits^0 \]
C. \[4500\,\mathop A\limits^0 \]
D. \[5000\,\mathop A\limits^0 \]
Answer
249.6k+ views
Hint:The least amount of energy necessary to cause photoemission of electrons from a metal surface is known as the work function. The maximum wavelength that incoming light must have for the photoelectric effect to occur is known as the threshold wavelength. The threshold wavelength will depend on the work function of the material and it does not depend on the intensity of incident radiation.
Formula used:
Work function is given as:
\[{W_0} = h\upsilon = \dfrac{{hc}}{{{\lambda _0}}}\]
Where, h is Planck’s constant, c is the speed of light, \[\upsilon \] 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 of metal, \[{W_0} = 1.3\,{\rm{ eV}}\]
As we know that work function is,
\[{W_0} = h\upsilon = \dfrac{{hc}}{{{\lambda _0}}}\]
Here \[{\lambda _0}\] is the threshold frequency
\[{\lambda _0} = \dfrac{{hc}}{W}\]
Substituting the values, we get
\[{\lambda _0} = \dfrac{{12375\,\mathop A\limits^0 \cdot eV}}{{3\,eV}}\]
\[\therefore {\lambda _0} = 4125\,\mathop A\limits^0 \]
Therefore, the threshold wavelength will be, \[{\lambda _0} = 4125\,\mathop A\limits^0 \].
Hence option A is the correct answer.
Note: 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. The threshold frequency is known as the minimum wavelength that the light should have for the photoelectric effect. So if the wavelength of the light is below the frequency of the threshold level, there will be no photoelectric effect. The threshold wavelength is commonly known as the threshold frequency. The relation between the frequency and the wavelength of light is \[\upsilon = c\lambda \]. When the light wave has a threshold frequency, its wavelength is called threshold wavelength.
Formula used:
Work function is given as:
\[{W_0} = h\upsilon = \dfrac{{hc}}{{{\lambda _0}}}\]
Where, h is Planck’s constant, c is the speed of light, \[\upsilon \] 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 of metal, \[{W_0} = 1.3\,{\rm{ eV}}\]
As we know that work function is,
\[{W_0} = h\upsilon = \dfrac{{hc}}{{{\lambda _0}}}\]
Here \[{\lambda _0}\] is the threshold frequency
\[{\lambda _0} = \dfrac{{hc}}{W}\]
Substituting the values, we get
\[{\lambda _0} = \dfrac{{12375\,\mathop A\limits^0 \cdot eV}}{{3\,eV}}\]
\[\therefore {\lambda _0} = 4125\,\mathop A\limits^0 \]
Therefore, the threshold wavelength will be, \[{\lambda _0} = 4125\,\mathop A\limits^0 \].
Hence option A is the correct answer.
Note: 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. The threshold frequency is known as the minimum wavelength that the light should have for the photoelectric effect. So if the wavelength of the light is below the frequency of the threshold level, there will be no photoelectric effect. The threshold wavelength is commonly known as the threshold frequency. The relation between the frequency and the wavelength of light is \[\upsilon = c\lambda \]. When the light wave has a threshold frequency, its wavelength is called threshold wavelength.
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