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Question

Ultraviolet light of wavelengths 800oA and 700oA when allowed to fall on hydrogen atoms in their ground states is found to liberate electrons with kinetic energies 1.8 eV and 4.0 eV, respectively. If the value of Planck's constant is 6.5×10X [1 electrons volt (eV)=1.6×1019J]. Find X?

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Solution

The energy of incident photon
=(hc/λ)
If Wi is the ionization energy and Ek the kinetic energy of emitted electron, then we have
hcλ1=W1+EK1
Therefore, for incident photon of wavelength
λ1=800oA=8×108m
hcλ=W1+EK1 (i)
And for incident photon of wavelength
λ2=700oA=7×108m,
hcλ=W1+EK2 (ii)
Subtracting Eqs. (ii) from (i), we get
hc(1λ21λ1)=EK2EK1
or h=(EK2EK1)λ1λ2c(λ1λ2)
Here, EK1=1.eV=1.8×1.6×1019J
and EK2=4.0eV=4.0×1.6×1019J
Substituting given values, we get
h=(4.81.8)×1.6×1019×8×108×7×1083×108(8×1087×108)
=6.57×1034Js

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