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Question: When a metallic surface is illuminated with light of wavelength \(\lambda \), the stopping potential...

When a metallic surface is illuminated with light of wavelength λ\lambda , the stopping potential is x volt. When the same surface is illuminated by light of wavelength 2λ2\lambda , the stopping potential is x3\dfrac{x}{3}. Threshold wavelength for the metallic surface is:
A) 4λ3\dfrac{{4\lambda }}{3}.
B) 4λ4\lambda .
C) 6λ6\lambda .
D) 8λ3\dfrac{{8\lambda }}{3}.

Explanation

Solution

Stopping voltage is defined as the potential difference that is applied to stop the moving electron from the late and generating current. Threshold frequency is the minimum frequency that must be given in the wave so as to make the flow of electrons from the plate.

Formula used: The formula of the Einstein equation for photoelectric effect is given by,
Voehc=1λ1λo\Rightarrow \dfrac{{{V_o}e}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}
Where threshold potential difference is Vo{V_o} the threshold wavelength is λo{\lambda _o} the wavelength of the wave is λ\lambda the charge of the electron is e the Planck’s constant is h and the speed of light is c.

Complete step by step solution:
It is given in the problem that when a metallic surface is illuminated with light of wavelength λ\lambda , the stopping potential is x volt and the same surface is illuminated by light of wavelength 2λ2\lambda , the stopping potential is x3\dfrac{x}{3} and we need to find the threshold wavelength for the metallic surface.
For the metallic surface with wavelength of λ\lambda .
The formula of the Einstein equation for photoelectric effect is given by,
Voehc=1λ1λo\Rightarrow \dfrac{{{V_o}e}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}
Where threshold potential difference is Vo{V_o} the threshold wavelength is λo{\lambda _o} the wavelength of the wave is λ\lambda the charge of the electron is e the Planck’s constant is h and the speed of light is c.
The stopping voltage is x and the wavelength is λ\lambda ,
Voehc=1λ1λo\Rightarrow \dfrac{{{V_o}e}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}
xehc=1λ1λo\Rightarrow \dfrac{{x \cdot e}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}………eq. (1)
For the wavelength of stopping voltage of x3\dfrac{x}{3} wavelength 2λ2\lambda .
The formula of the Einstein equation for photoelectric effect is given by,
Voehc=1λ1λo\Rightarrow \dfrac{{{V_o}e}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}
Where threshold potential difference is Vo{V_o} the threshold wavelength is λo{\lambda _o} the wavelength of the wave is λ\lambda the charge of the electron is e the Planck’s constant is h and the speed of light is c.
Voehc=1λ1λo\Rightarrow \dfrac{{{V_o}e}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}
xe3(hc)=12λ1λo\Rightarrow \dfrac{{x \cdot e}}{{3\left( {h \cdot c} \right)}} = \dfrac{1}{{2\lambda }} - \dfrac{1}{{{\lambda _o}}}………eq. (2)
Taking ratio of equation (1) and equation (2).
(xehc)(xe3(hc))=(1λ1λo)(12λ1λo)\Rightarrow \dfrac{{\left( {\dfrac{{x \cdot e}}{{hc}}} \right)}}{{\left( {\dfrac{{x \cdot e}}{{3\left( {h \cdot c} \right)}}} \right)}} = \dfrac{{\left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}} \right)}}{{\left( {\dfrac{1}{{2\lambda }} - \dfrac{1}{{{\lambda _o}}}} \right)}}
3=(1λ1λo)(12λ1λo)\Rightarrow 3 = \dfrac{{\left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}} \right)}}{{\left( {\dfrac{1}{{2\lambda }} - \dfrac{1}{{{\lambda _o}}}} \right)}}
3(12λ1λo)=(1λ1λo)\Rightarrow 3\left( {\dfrac{1}{{2\lambda }} - \dfrac{1}{{{\lambda _o}}}} \right) = \left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}} \right)
32λ3λo=1λ1λo\Rightarrow \dfrac{3}{{2\lambda }} - \dfrac{3}{{{\lambda _o}}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}
32λ=1λ+2λo\Rightarrow \dfrac{3}{{2\lambda }} = \dfrac{1}{\lambda } + \dfrac{2}{{{\lambda _o}}}
32λ1λ=2λo\Rightarrow \dfrac{3}{{2\lambda }} - \dfrac{1}{\lambda } = \dfrac{2}{{{\lambda _o}}}
322λ=2λo\Rightarrow \dfrac{{3 - 2}}{{2\lambda }} = \dfrac{2}{{{\lambda _o}}}
12λ=2λo\Rightarrow \dfrac{1}{{2\lambda }} = \dfrac{2}{{{\lambda _o}}}
λo=4λ\Rightarrow {\lambda _o} = 4\lambda.
The threshold frequency is equal to λo=4λ{\lambda _o} = 4\lambda .

The correct answer for this problem is option B.

Note: It is advisable for students to understand and remember the formula of the photoelectric effect as it is helpful in solving problems like these. The threshold frequency is the minimum frequency which is required to make the electrons flow.