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Light of wavelength 5500 Å falls on a sensitive plate with photoelectric work function of 2 eV. The kinetic energy of the photoelectrons emitted will be in eV____.
[Planck’s constant, h=6.63×10^–34 Js]
[Round off to two decimal places]

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answer is 0.26.

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Detailed Solution

$\begin{array}{l} h ν = W_{0} + K \cdot E_{max} \\ E = W_{0} + K \cdot E_{max} \\ ∴ K \cdot E_{max} = E - W_{0} \\ = \frac{h c}{λ e} - 2 = 2.26 - 2 \\ K \cdot E_{max} = 0.26 eV \end{array}$

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