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The photons from the Balmer series in Hydrogen spectrum having wavelength between `450 nm` to `700 nm` are incident on a metal surface of work function `2 eV` Find the maximum kinetic energy of ejected electron (Given hc = 1242 eV nm)` |
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Answer» Balmer series `lambda_(32) = (12375)/(E_3 - E_2) = (12375)/((13.6)((1)/(4)- (1)/(9))` =6551 Å = 655.1 nm `lambda_(42) = (12375)/(E_4 -E_2) = (12375)/((13.6)((1)/(4) - (1)/(16)))` = 4853 Å =433.3 nm First two lie in the given range. of these `lambda_(42)` corresponds to more energy `E = E_4 - E_2 = (13.6)((1)/(4) - (1)/(16))` =2.55 eV `K_(max) = E -W = (2.55 - 2.0) eV` = 0.55 eV. |
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