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प्रश्न
The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which one of the two has higher value of work-function? Justify your answer.
The graph shows variation of stopping potential V0 versus frequency of incident radiation v for two photosensitive metals A and B. Which of the two metals has higher threshold frequency and why?
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उत्तर
eV0=hν−hν0
`V_0-h/ev-h/ev_0`
Here, e is the charge of an electron, Vo is the stopping potential, h is Planck's constant, ν0 is the threshold frequency and ν is the frequency of the incident light.
On comparing with y=mx+c, we find that the more the magnitude of intercept on y or V0 axis, the more is the threshold frequency (ν0).
From the graph, the threshold frequency for metal A is greater than that for metal B. Hence, the work function for metal A is greater than that for metal B.
संबंधित प्रश्न
The photoelectric work function for a metal surface is 2.3 eV. If the light of wavelength 6800A is incident on the surface of metal, find threshold frequency and incident frequency. Will there be an emission of photoelectrons or not?
[Velocity of light c = 3 x 108 m/s,
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