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When light of wavelength lambda is incid...

When light of wavelength `lambda` is incident on a metal surface, stopping potential is found to be x. When light of wavelength n`lambda` is incident on the same metal surface, stopping potential is found to be `(x)/(n+1)` . Find the threshold wavelength of the metal.

Text Solution

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Let `lambda_(0)` is the threshold wavelength. The work function is `phi = (hc)/(lambda_(0))`
Now, by photoelectric equation `ex = (hc)/(lambda) - (hc)/(lambda_(0))` ……(i) `(ex)/(n+1) = (hc)/(nlambda) - (hc)/(lambda_(0))`……..(ii)
From (i) and (ii) `(hc)/(lambda) - (hc)/(lambda_(0)) = (n+1) (hc)/(nlambda) - (n+1) (hc)/(lambda_(0)) rArr (nhc)/(lambda_(0)) = (hc)/(nlambda) rArr lambda_(0) = n^(2)lambda`
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