1.

The graph between the stopping potential `V_(0)` and frequency `v` for two different metal plates `P` and `Q` are shown in the figure. Which of the metals has greater threshold wavelength and work function.

Answer» `eV_(s)= hv-W`
`W=(hc)/(lambda_(th))`
From graph,
`W_(Q) gt W_(p)` So work function of `Q` is greater than `P`
`lambda_(th p) gt lambda_(th Q)`


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