The work function of caesium metal is 2.14 eV. When light of frequency 6 x 10¹⁴Hz is incident on the metal surface, photoemission of electrons occurs. What is the maximum speed of the emitted photoelectrons?

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Solution

Given, a ceasium metal.
Work function = 2.14 eV
Frequency of light,

ν

= 6 x 10¹⁴Hz

Maximum kinetic energy is given by,

\frac{1}{2} {mv}^{2}_{\max} = 0.346 eV = 0.346 \times 1.6 \times 10^{- 19} J

\Rightarrow

v_{\max} = \sqrt{\frac{0.346 \times 1.6 \times 10^{- 19} \times 2}{9.1 \times 10^{- 31}}} {ms}^{- 1}

= 3.488 \times 10^{5} {ms}^{- 1} = 348.8 km s^{- 1}

i.e.,

v_{\max} = 349 km s^{- 1} .

which is the required maximum speed of the photoelectrons.

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