Question

What is the work function of the metal if the light of wavelength 4000 A° generates photoelectron of velocity 6 × 10⁵ m/s from it ?

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Answer 1

$\texttt{\textsf{\huge{\underline{Solution}:}}}$

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$KE = \frac{1}{2} \times 9 \times {10}^{ - 31} \times (6 \times {10}^{5} )^{2}$

$= 162 \times {10}^{ - 21} J≃eV$

$Energy \:\: of\: photon \: E = \frac{12400}{4000} = 3.1eV$

$E = ϕ + KE$

$3.1 =ϕ +1$

$ϕ= 2.1 eV$

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♥︎If there is any mistake in this answer please forgive me.☺︎︎

Answer 2

$\huge\color{blue}\boxed{\colorbox{lightgreen}{♧Given♧}}\bigstar$

$Mass \: of \: electron = 9 \times 10 {}^{ - 31} kg$

$Velocity \: of \: light = 3 \times 10 {}^{8} mg {}^{ - 1}$

$Planck \: constant = 6.626 \times 10 {}^{ - 34} Js$

$Charge \: election = 1.6 \times 10 {}^{ - 19} JeV {}^{ - 1}$

SOLUTION

Hv=ϕ+hv°

$\frac{1}{2} mv {}^{2} + hc( \frac{1}{ λ} - \frac{1}{λ})$

$Hv=ϕ+ \frac{1}{2} mv {}^{2}$

$ϕ = \frac{6.626 \times 10 { { {}^{34} \times 3 \times 10 {}^{8} }^{} } }{4000 \times 10 {}^{ - 10} } - \frac{1}{2} \times 9 \times 10 {} {}^{ - 31} \times(6 \times 10 {}^{3} ) {}^{2}$

$ϕ = 3.35 \times 10 {}^{ - 10} J$

$ϕ = 2.1 ev$

${\mathfrak{\purple{Thanks}}}$.