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Svetach [21]
3 years ago
11

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 ✕ 10−19 J, e = 1.602 ✕ 10−19 C, c = 2.998 ✕ 108 m/s, and h = 6.626 ✕ 10−34 J · s = 4.136 ✕ 10−15 eV · s as necessary.)
Physics
1 answer:
bulgar [2K]3 years ago
3 0

Answer:\lambda =248.99 nm

Explanation:

Given

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

h=6.626\times 10^{-34} J

c=2.998\times 10^8

\phi =\frac{hc}{\lambda }

\lambda =\frac{hc}{\phi }

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

\lambda =248.99 nm

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F=18.58\times 10^{-11}\ N

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<em>For angle in counterclockwise direction from the +x-axis</em>

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