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Q.When a light of frequency 9×10149 \times 10^{14} Hz is incident on a metal surface, photoelectrons are emitted with a maximum speed of 8.14×105 ms−18.14 \times 10^{5}\ \text{ms}^{-1}. Determine the threshold frequency of the surface.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2026Subjective· 3mImportance★★★★★
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Computing the photoelectrons' maximum kinetic energy from their given speed, then applying Einstein's photoelectric equation, gives a threshold frequency of about 4.45×10144.45\times10^{14} Hz.

Working

1. Maximum kinetic energy of the photoelectrons.

KEmax=12mv2=12(9.11×10−31)(8.14×105)2KE_{max} = \dfrac12mv^2 = \dfrac12(9.11\times10^{-31})(8.14\times10^5)^2

=12(9.11×10−31)(6.626×1011)≈3.02×10−19 J= \dfrac12(9.11\times10^{-31})(6.626\times10^{11}) \approx 3.02\times10^{-19}\ \text{J}

2. Einstein's photoelectric equation.

hν=hu0+KEmax ⇒ u0=ν−KEmaxhh\nu = h u_0 + KE_{max} \ \Rightarrow\ u_0 = \nu - \dfrac{KE_{max}}{h}

3. Substituting (ν=9×1014\nu=9\times10^{14} Hz, h=6.63×10−34h=6.63\times10^{-34} Js): …

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