Question:
When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0 \,V$. This potential drops to $0.6 \,V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? \([ Take=\frac{h c}{e}=124 \times 10^{-6} Jm C ^{-1}]\)
When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0 \,V$. This potential drops to $0.6 \,V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? \([ Take=\frac{h c}{e}=124 \times 10^{-6} Jm C ^{-1}]\)
Updated On: May 25, 2024
- $1.72 \times 10^{-7} m , 1.20 \, eV$
- $1.72 \times 10^{-7} m , 5.60 \,eV$
- $3.78 \times 10^{-7} m , 5.60 \,eV$
- $3.78 \times 10^{-7} m , 1.20 \, eV$
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The Correct Option is A
Solution and Explanation
The correct answer is $1.72 \times 10^{-7} m , 1.20 \, eV$ ;opton (A).
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The least energy that is needed to emit an electron from the surface of a metal can be supplied to the loose electrons.
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The photoelectric effect is a phenomenon that involves electrons getting away from the surface of materials.
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