Given that a photon of light of wavelength 1000 Å has an energy equal to 1.23 eV. When light of wavelength 5000 Å and intensity I0 falls on photoelectric cell, the saturation current is 0 . 40 x 106 amp. and the stopping potential is 1.36 V. Then the work-function is

A
0.43 eV
B
1.10 eV
C
1.36 eV
D
2.47 eV

Solution:

$hv = W + {eV}_{0} where e V_{o} = sto p p ing Potential W = hv - {eV}_{0} Incident wavelength = 5000 Å hv = energy of 5000 Å = 2 \times 1 \cdot 23 eV = 2 \cdot 46 eV {eV}_{0} = 1 \cdot 36 eV W = (2 \cdot 46 - 1 \cdot 36) = 1 \cdot 10 eV$

Given that a photon of light of wavelength 1000 $Å$ has an energy equal to 1.23 eV. When light of wavelength 5000 $Å$ and intensity I₀ falls on photoelectric cell, the saturation current is 0 . 40 x 10⁶amp. and the stopping potential is 1.36 V. Then the work-function is

Solution:

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