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The critical wavelength for producing photoelectric effect in a metal is `2500 Å` What wavelength would be nuccesary be produce photoelectric effect from this metal , having twice the KE of these produced at `2000 Å` |
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Answer» `KE_(2000) = hv_(2000) - hv_(0)= hc ((1)/(lambda_(2000)) - (1)/(lambda_(0)))` `KE_("new") - hv_("new") = hv_(0)= hc ((1)/(lambda_("new")) - (1)/(lambda_(0)))` KE_("new") = 2 xx KE_(2000)` `2= (hc((1)/(lambda_("new")) - (1)/(2500)))/(hc((1)/(2000) - (1)/(2500))) , lambda = 1666.66 Å` |
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