In a photoelectric effect experiment, the stopping potential for a metal surface is 2.0 V when light of wavelength 300 nm is used. What is the work function of the metal?

B. 2.13 eV — Using Einstein's photoelectric equation, the energy of the incident photon is E = hc/lambda = 1240/300 = 4.13 eV. The work function is Phi = E - eV_s = 4.13 - 2.0 = 2.13 eV.

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$In a photoelectric effect experiment, the stopping potential for a metal surface is V_{s} = 2.0 V when light of wavelength λ = 300 nm is used. What is the work function Φ of the metal? (Take h c = 1240 eV nm)$

A. $1.13 eV$
B. $2.13 eV$
C. $3.13 eV$
D. $4.13 eV$

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Correct answer: B — $2.13 eV$

Solution

Using Einstein's photoelectric equation, the energy of the incident photon is E = h c / λ = 1240/300 = 4.13 eV. The work function is Φ = E - e V_{s} = 4.13 - 2.0 = 2.13 eV.

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In a photoelectric effect experiment, the stopping potential for a metal surface is Vs​=2.0 V when light of wavelength λ=300 nm is used. What is the work function Φ of the metal? (Take hc=1240 eV nm)

Solution

Attempt this question & track your score

More Modern Physics practice questions

$In a photoelectric effect experiment, the stopping potential for a metal surface is V_{s} = 2.0 V when light of wavelength λ = 300 nm is used. What is the work function Φ of the metal? (Take h c = 1240 eV nm)$