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3 years ago

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A magnesium surface has a work function of 3.60 eV. Electromagnetic waves with a wavelength of 320 nm strike the surface and eje

ct electrons. Find the maximum kinetic energy of the ejected electrons. Express your answer in electron volts.

1 answer:

Vladimir79 [104]3 years ago

6 0

Answer:

K(maximum) = 0.284 eV

Explanation:

Given data

work function ω = 3.60 eV = 3.60 × 1.6 × $10^{-19}$ = 5.76 × $10^{-19}$ J

wavelength λ = 320 nm = 320 × $10^{-9}$ m

to find out

maximum kinetic energy

solution

we know that speed of light is = 3 × $10^{8}$ m/s

and planck constant h = 6.63 × $10^{-34}$ J.s

we will apply here Einstein's photo elecric equation that is

hc/λ = ω + K(maximum)

put here all these value we get

6.63 × $10^{-34}$ × 3 × $10^{8}$ / 320 × $10^{-9}$ = 5.76 × $10^{-19}$ + K(maximum)

so

K(maximum) = 6.63 × $10^{-34}$ × 3 × $10^{8}$ / 320 × $10^{-9}$ - 5.76 × $10^{-19}$

K(maximum) = 0.4556 × $10^{-19}$ J

K(maximum) = 0.4556 × $10^{-19}$ / 1.6 × $10^{-19}$

K(maximum) = 0.284 eV

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The final temperature is $T_f = 78 ^o F = (78-32) *\frac{5}{9} +273.15 = 298.7 \ K$

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Generally the quantity of heat energy needed is mathematically represented as

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=> $Q = 51250 * 0.099 *10^3 * (298.7 - 152)$

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