Calculating W 0 W_0 of metal

Chemistry Level 2

A light source of wavelength λ \lambda illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1.00 eV. A second light source of wavelength λ 2 \frac{\lambda}{2} shines on the same metal and ejects photoelectrons with a maximum kinetic energy of 4.00 eV. Calculate the work function of the metal.

3.2043e-20 J 2 eV 5 eV 0.000124 ev 1.6022e-20 J

Winod Dhamnekar by Winod DHAMNEKAR , 55, India

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2 solutions

E1 = Ke1 +W (a)

E2 = Ke2 + W (b)

E2 = Ke2 + (E1-Ke1)

hc/lambda2 = Ke2 + (hc/lambda1 -Ke1)

Is known that lambda2=lambda1/2, so:

2hc/lambda1 = ke2 + (hc/lambda1 - Ke1)

2hc=[ke2 + (hc/lambda1 - Ke1)] × lambda1

2hc = lambda1×Ke2 + hc - lambda1×Ke1

2hc-hc = lambda1(ke2-ke1)

lambda1 = hc/ke2 - ke1

lambda1 = 1.988E-25/(6.4E-19 - 1.6E-19)

lambda1 = 414 mm.

Substitute lambda 1 in (a):

W =[ (6.6E-34 × 3.0E8) / 414E-9 ] - 1.6E-19 = 3.2E-19 J ( ~ 2.0 eV).

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@Tibrôncio Cavalcante , Please learn formatting guide available on https://brilliant.org, so that other readers can comprehend and read it easily.

Winod DHAMNEKAR - 12 months ago

h c λ = W 0 + 1.00 \dfrac {hc}{\lambda}=W_0+1.00 where h h is Planck's constant and c c is the speed of light in vacuum.

2 h c λ = W 0 + 4.00 \dfrac{2hc}{\lambda}=W_0+4.00

So W 0 + 4.00 = 2 W 0 + 2.00 ⟹ W 0 = 2 W_0+4.00=2W_0+2.00\implies W_0=\boxed 2 eV.

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