Find the maximum velocity of photoelectrons emitted by radiation of frequency 3 x 10¹⁵ Hz from a photoelectric surface having a work function 4.0 eV.

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Solution

Given,
Frequency of radiation,

ν

= 3 x 10¹⁵ Hz
Work function of the material,

ϕ_{o}

= 4.0 eV

As per Einstein's photoelectric equation,

\frac{1}{2} {mv}^{2}_{\max} = hv - ϕ_{0}

= 6.63 \times 10^{- 34} \times 3 \times 10^{15} - 4 \times 1.6 \times 10^{- 19}

\Rightarrow

{v^{2}}_{\max} = \frac{2 [19.89 \times 10^{- 19} - 6.4 \times 10^{- 19}]}{9.1 \times 10^{- 31}}

= \frac{26.98 \times 10^{- 19}}{9.1 \times 10^{- 31}} = 2.96 \times 10^{12}

v_{\max} = 1.72 \times 10^{6} {ms}^{- 1} .

is the required maximum velocity of photoelectrons.

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