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When a metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential is 5 V0. When the same surface is illuminated with light of wavelength 3λ, the stopping potential is V0. Then the work function of the metallic surface is :
- a)hc/6λ
- b)hc/5λ
- c)hc/4λ
- d)2hc/4λ
Correct answer is option 'A'. Can you explain this answer?
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Explanation:
Given Data:
- Stopping potential for wavelength λ = 5V0
- Stopping potential for wavelength 3λ = V0
Formula:
The stopping potential is given by the equation:
\[ V_0 = \frac{hc}{\lambda} - \phi \]
Where:
- V0 is the stopping potential
- h is the Planck's constant
- c is the speed of light
- λ is the wavelength of the incident light
- φ is the work function of the metal
Calculations:
1. For wavelength λ:
\[ V_0 = \frac{hc}{\lambda} - \phi \]
\[ 5V_0 = \frac{hc}{\lambda} - \phi \] (1)
2. For wavelength 3λ:
\[ V_0 = \frac{hc}{3\lambda} - \phi \]
\[ V_0 = \frac{hc}{3\lambda} - \phi \] (2)
3. Subtracting equation (2) from equation (1):
\[ 5V_0 - V_0 = \frac{hc}{\lambda} - \frac{hc}{3\lambda} \]
\[ 4V_0 = \frac{2hc}{3\lambda} \]
\[ V_0 = \frac{hc}{6\lambda} \]
Therefore, the work function of the metallic surface is:
\[ \phi = hc/\lambda - V_0 = hc/\lambda - \frac{hc}{6\lambda} = \frac{hc}{6\lambda} \]
So, the correct answer is option 'A' (hc/6λ).
Given Data:
- Stopping potential for wavelength λ = 5V0
- Stopping potential for wavelength 3λ = V0
Formula:
The stopping potential is given by the equation:
\[ V_0 = \frac{hc}{\lambda} - \phi \]
Where:
- V0 is the stopping potential
- h is the Planck's constant
- c is the speed of light
- λ is the wavelength of the incident light
- φ is the work function of the metal
Calculations:
1. For wavelength λ:
\[ V_0 = \frac{hc}{\lambda} - \phi \]
\[ 5V_0 = \frac{hc}{\lambda} - \phi \] (1)
2. For wavelength 3λ:
\[ V_0 = \frac{hc}{3\lambda} - \phi \]
\[ V_0 = \frac{hc}{3\lambda} - \phi \] (2)
3. Subtracting equation (2) from equation (1):
\[ 5V_0 - V_0 = \frac{hc}{\lambda} - \frac{hc}{3\lambda} \]
\[ 4V_0 = \frac{2hc}{3\lambda} \]
\[ V_0 = \frac{hc}{6\lambda} \]
Therefore, the work function of the metallic surface is:
\[ \phi = hc/\lambda - V_0 = hc/\lambda - \frac{hc}{6\lambda} = \frac{hc}{6\lambda} \]
So, the correct answer is option 'A' (hc/6λ).
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When a metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential is 5 V0. When the same surface is illuminated with light of wavelength 3λ, the stopping potential is V0. Then the work function of the metallic surface is :a)hc/6λb)hc/5λc)hc/4λd)2hc/4λCorrect answer is option 'A'. Can you explain this answer?
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When a metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential is 5 V0. When the same surface is illuminated with light of wavelength 3λ, the stopping potential is V0. Then the work function of the metallic surface is :a)hc/6λb)hc/5λc)hc/4λd)2hc/4λCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When a metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential is 5 V0. When the same surface is illuminated with light of wavelength 3λ, the stopping potential is V0. Then the work function of the metallic surface is :a)hc/6λb)hc/5λc)hc/4λd)2hc/4λCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When a metallic surface is illuminated with monochromatic light of wavelength λ, the stopping potential is 5 V0. When the same surface is illuminated with light of wavelength 3λ, the stopping potential is V0. Then the work function of the metallic surface is :a)hc/6λb)hc/5λc)hc/4λd)2hc/4λCorrect answer is option 'A'. Can you explain this answer?.
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