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Light form a discharge tube containing `H` atoms falls on the sodium metal surface. The kinetic energy of the fastest moving photoelectron emitted from sodium is `0.73 eV`. If these photons are emitted in `H-`atom due to the transition from engergy level `(n_(2))` to `(n_(1))` and the work function of sodium metal is `1.82 eV`. then the minimum value of `(n_(1) + n_(2))` is .........

Answer» Correct Answer - `6`
Energy of emitted photon `= 1.82 + 0.73 = 2.55 eV`
So, `DeltaE = 13.6 xx 1^(2)[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` for `n_(1) = 2` & `n_(2) = 4`


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