• According to ,quantum theory:
hv =hv0 + 1/2m v2
1/2m v2 = h(v-v0)
where h = plank’s constant = 6.626 x 10-34 J S,
v0 = threshold frequency of photons , which we have to calculate
v = frequency of incident photons. Given = 1.0×1015 s-1
1/2m v2 = kinetic energy of electrons given = 1.988 × 10-19 J
= 1.0×1015 s-1 - 0.300030 × 1015
= 6.9997 × 10 14 s-1 (threshold frequency)
• The relation b/w wavelength and frequency , v = c/λ , where λ = wavelength and c= velocity of light = 3 × 108 m/s ,
• Here in case of threshold frequency v0 = c/λ0
Hence the maximum wavelength of the photon will be
4.36 × 10-7 m which is definitely greater than
Now, 600 nm = 6 × 10-7m is definitely greater than the calculated λ0 value.
hence, the electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.
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