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Light from a discharge tube containing H-atoms in some excited state, falls on the metallic surface of metal Na.The KE of the fastest photo-electron was found to be fastest photo-electron was found to be `10.93eV`. If `He^(+)` ions were present in the same excited state, the KE of the fastest photo-electron would have been `49.18eV`. Determine the excited state orbit number and work function of Na.A. `2,18.2ev`B. `4,1.82ev`C. `3,16ev`D. `2,1.6ev` |
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Answer» Correct Answer - B `E_(1) = E_(0) +10.93, E_(2) = E_(0) +49.18` `[13.6 xx 2^(2) - 13.6 xx 1^(2)] [(1)/(1^(2))-(1)/(n^(2))]` `= E_(2) - E_(1) = Delta E = 38.25` `:. n = 4, :. E_(1) = 12.75 & E_(0) = 1.82 eV` |
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