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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
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A beam of monochromatic radiation is incident on a photosensitive surface. Answer the following question giving reason :
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If the frequency of the incident radiation is increased from 4 × 1015 Hz to 8 × 1015 Hz, by how much will the stopping potential for a given photosensitive surface go up?
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Where are the electrons that absorb energy from incident radiation located?
What must electrons near the metal surface overcome before they can escape?
