The work function for a metal surface is 4.98 eV. What is the largest wavelength of light in nm that will produce photoelectrons from this surface? (Use 1 eV = 1.602 βœ• 10βˆ’19 J, e = 1.602 βœ• 10βˆ’19 C, c = 2.998 βœ• 108 m/s, and h = 6.626 βœ• 10βˆ’34 J Β· s = 4.136 βœ• 10βˆ’15 eV Β· s as necessary.)

Respuesta :

Answer:[tex]\lambda =248.99 nm[/tex]

Explanation:

Given

Work function[tex]\left ( \phi \right )=4.98\approx 1.602\times 10^{-19}\times 4.98[/tex]

[tex]h=6.626\times 10^{-34} J[/tex]

[tex]c=2.998\times 10^8[/tex]

[tex]\phi =\frac{hc}{\lambda }[/tex]

[tex]\lambda =\frac{hc}{\phi }[/tex]

[tex]\lambda =\frac{6.626\times 10^{-34}\times 2.998\times 10^8}{4.98\times 1.602\times 10^{-19}} [/tex]

[tex]\lambda =248.99 nm[/tex]

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