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Q.

Photons of energy  2eV fall on a metal plate and release photo electrons with a maximum velocity v , by decreasing the wave length by 25% the maximum velocity of photo electrons is doubled. The work function of metal of the material of plate in eV  is

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answer is 0001.78.

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Detailed Solution

v1=v,v2=2vWe know that12mv2=hcλ−w           here w=work function∴v12v22=12400λ1−w12400λ2−wGiven : energy of photon=12400λ1=2eVNew wavelength=λ2=34λ1New energy of photon=12400λ2=1240034λ1=43(2)eV⇒v24v2=2−w43(2)−w⇒8−4w=83−w⇒8−83=4w−w⇒163=3w⇒w=169=1.78eV
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