
When a light of frequency ${v_1}$ is incident on a metal surface, the photoelectrons emitted have twice the kinetic energy as did the photoelectrons emitted when the same metal has irradiated with light of frequency ${v_2}$. What will be the value of threshold frequency$?$
$
A){v_0} = {v_1} - {v_2} \\
B){v_0} = {v_1} - 2{v_2} \\
C){v_0} = 2{v_2} - {v_1} \\
D){v_0} = {v_1} + {v_2} \\
$
Answer
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Hint: Threshold energy is the minimum frequency of radiation that will produce a photoelectric effect below which the photoelectric emission is not possible completely. Irrespective of the intensity of incident radiation. Light at the threshold frequency will dislodge the electron with no kinetic energy.
Complete answer: The photoelectric threshold frequency, symbolized by the Greek letter nu with subscript zero, ${v_0}$, is that frequency at which the effect is barely possible, it is given by the ratio of the work function symbolized by the Greek letter psi, $\psi $, to Planck’s constant.
Threshold energy is the amount of energy it takes to remove an electron from a metallic surface, just like the work function.
$ \Rightarrow KE = \dfrac{1}{2}m{v^2}$
We know $E = h{v_0}$,
So, we can write it as for frequency${v_1}$:
$ \Rightarrow h{v_1} = h{v_0} + K{E_1}$, consider it as equation $(1)$
For, frequency${v_2}$:
$ \Rightarrow h{v_2} = h{v_0} + K{E_2}$, consider it as equation $(2)$
It is given that$K{E_1} = 2K{E_2}$, equation $(3)$
Now, substitute equation $(3)$ in equation $(1)$
$ \Rightarrow h{v_1} = h{v_0} + 2K{E_2}$, equation$(4)$
Form equation $(2)$
$ \Rightarrow h{v_2} - h{v_0} = K{E_2}$, equation $(5)$
Now substitute equation $(5)$in equation $(4)$
$ \Rightarrow h{v_1} = h{v_0} + 2(h{v_2} - h{v_0})$
$ \Rightarrow h{v_1} = h{v_0} + 2h{v_2} - 2h{v_0}$
$ \Rightarrow h{v_1} = 2h{v_2} - h{v_0}$
$ \Rightarrow h{v_0} = 2h{v_2} - h{v_1}$
$ \Rightarrow {v_0} = 2{v_2} - {v_1}$
So, the correct answer is $C){v_0} = 2{v_2} - {v_1}$.
Note:
Remember threshold frequency should not be confused with the threshold wavelength. Threshold frequency is the minimum frequency of the light required to remove an electron out of a metal and threshold wavelength is the maximum wavelength that the incident radiation must have so that photoelectric effect takes place.
Complete answer: The photoelectric threshold frequency, symbolized by the Greek letter nu with subscript zero, ${v_0}$, is that frequency at which the effect is barely possible, it is given by the ratio of the work function symbolized by the Greek letter psi, $\psi $, to Planck’s constant.
Threshold energy is the amount of energy it takes to remove an electron from a metallic surface, just like the work function.
$ \Rightarrow KE = \dfrac{1}{2}m{v^2}$
We know $E = h{v_0}$,
So, we can write it as for frequency${v_1}$:
$ \Rightarrow h{v_1} = h{v_0} + K{E_1}$, consider it as equation $(1)$
For, frequency${v_2}$:
$ \Rightarrow h{v_2} = h{v_0} + K{E_2}$, consider it as equation $(2)$
It is given that$K{E_1} = 2K{E_2}$, equation $(3)$
Now, substitute equation $(3)$ in equation $(1)$
$ \Rightarrow h{v_1} = h{v_0} + 2K{E_2}$, equation$(4)$
Form equation $(2)$
$ \Rightarrow h{v_2} - h{v_0} = K{E_2}$, equation $(5)$
Now substitute equation $(5)$in equation $(4)$
$ \Rightarrow h{v_1} = h{v_0} + 2(h{v_2} - h{v_0})$
$ \Rightarrow h{v_1} = h{v_0} + 2h{v_2} - 2h{v_0}$
$ \Rightarrow h{v_1} = 2h{v_2} - h{v_0}$
$ \Rightarrow h{v_0} = 2h{v_2} - h{v_1}$
$ \Rightarrow {v_0} = 2{v_2} - {v_1}$
So, the correct answer is $C){v_0} = 2{v_2} - {v_1}$.
Note:
Remember threshold frequency should not be confused with the threshold wavelength. Threshold frequency is the minimum frequency of the light required to remove an electron out of a metal and threshold wavelength is the maximum wavelength that the incident radiation must have so that photoelectric effect takes place.
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