When hydrogen atom absorbs a photon of wavelength 60 nm, photo-ionization of atom occurs. The maximum kinetic energy of emitted electron will be

A
13.6 eV
B
7'15 eV
C
3'6 eV
D
1'9 eV

Solution:

The energy of photon of wavelength of 60 nm will be
$= \frac{hc}{λ} = \frac{(6 \cdot 64 \times 10^{- 34}) (3 \times 10^{8})}{(60 \times 10^{- 9}) (1 \cdot 6 \times 10^{- 9})} eV = 20 \cdot 75 eV We know that binding energy of electron in ll - atom = 13.6 eV . K . E . of emitted electron = zo . 7 s - t 3 - 6 = 7.15 eV$

Solution:

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