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in Dual Nature of Radiation and Matter by (26.7k points)
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The work function of caesium metal is 2.14 eV. When light of frequency 6 × 1014 Hz is incident on the metal surface, photo emission of electrons occurs. What is the

1. maximum kinetic energy of the emitted electrons,

2. maximum speed of the emitted photo electrons?

1 Answer

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Best answer

Given W0 = 2.14 eV
= 2.14 × 1.6 × 10-19 J
= 3.424 × 10-19J
v = 6 × 1014 Hz

1. Kmax = hv – W0
= 6.62 × 10-34 × 6 × 1014 – 3.424 × 1019
= 3.972 × 10-19 – 3.424 × 10-19
= 0.54 × 10-19 J.

2. Since eV0 =kmax

∴ v0 = \(\frac{k_{max}}{e}\) \(\frac{0.54\times10^{-19}}{1.6\times10^{-19}}\)

or v0 = 0.34 V

Since \(\frac{1}{2}\) mV2max = Kmax

∴ V2max = \(\frac{2k_{max}}{m}\)

\(\frac{2\times0.54\times10^{-19}}{9.1\times10^{-31}}\) = 0.1186

or vmax = 0.344 × 106 ms-1 = 344 kms-1

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