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Hydrogen gas in the atomic state is exci...

Hydrogen gas in the atomic state is excited to an energy level such that the electrostatic potential energy of H-atom becomes `-1.7 eV`. Now, a photoelectric plate having work function w=2.3 eV is exposed to the emission spectra of this gas. Assuming all the transitions to be possible, find the minimum de-Broglie wavelength of the ejected photoelectrons.

Text Solution

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Given that electrostatic potential energy `= - 1.7 e V`
`:. KE = (1.7)/(2) = 0.85 e V`
So, total energy `= - 1.7 + 0.85 = - 0.85 e V`
Now, `E_(n) = - (13.6)/(n^(2)) = - 0.85`
`:. n^(2) = (13.6)/(0.85) or n = 4`
Hence the atom is excited to state `n = 4`
`Delta E = - 0.85 - (-13.6) = 12.75 e V`
Using Einstein's equation , we have
`hvm = W + (KE)_(max)`
or ` (KE)_(max) = hv - W = 12.75 - 2.3 = 10.45 e V`
The minimum de Broglie wavelength is given by
`lambda_(min) = (h)/((p) _(max)) = (h)/sqrt(2 m (KE)_(max))`
`= (6.63 xx 10^(_34))/(sqrt(2 xx (9.1 xx 10^(-31)) (10.45 xx 1.6 xx 10^(-19))))`
`= 3.8 xx 10^(-10) m = 3.8 Å`
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