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Paha777 [63]
3 years ago
5

What are the cues for a forehand stroke

Physics
1 answer:
MaRussiya [10]3 years ago
8 0

Answer:

huh

Explanation:

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In the photoelectric effect, the energy carried by the incoming photon is used partially to extract the photoelectron from the material  and the rest is converted into kinetic energy of the electron:
hf = \phi + K_{max} (1)
where
h is the Planck constant
f is the photon frequency
\phi is the work function of the material (the energy needed to extract the photoelectron)
K_{max} is the maximum kinetic energy of the photoelectron

The ultraviolet light has a wavelength of \lambda=400 nm=400 \cdot 10^{-9} m, so its frequency is
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{400 \cdot 10^{-9} m}=7.5 \cdot 10^{14}Hz
And the energy of each photon of this light is
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The maximum kinetic energy of the emitted photoelectrons is (converting into Joules)
K_{max} = 1.10 eV  \cdot 1.6 \cdot 10^{-19} J/eV =1.76 \cdot 10^{-19}J

And so by using equation (1) we can find the work function of the material:
\phi = hf - K_{max} = 4.95 \cdot 10^{-19}J - 1.76 \cdot 10^{-19} J =3.19 \cdot 10^{-19}J

Now we have instead light wavelength \lambda=270 nm= 270 \cdot 10^{-9}m hitting the same surface. The frequency of this light is
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{270 \cdot 10^{-9}m}=1.11 \cdot 10^{15} Hz
and the energy of each photon of this light is
E=hf=(6.6 \cdot 10^{-34}Js)(1.11 \cdot 10^{15}Hz)=7.33 \cdot 10^{-19} J

And so we can calculate the new maximum kinetic energy of the photoelectrons by using (1) and the work function we found previously:
K_{max} = hf-\phi = 7.33 \cdot 10^{-19}J - 3.19 \cdot 10^{-19}J = 4.14 \cdot 10^{-19}J

And if we want to convert it into electronvolts,
K_{max} =  \frac{4.14 \cdot 10^{-19}J}{1.6 \cdot 10^{-19} J/eV}=2.59 eV 

The maximum kinetic energy of the photoelectrons has increased because the light hitting the surface is now more energetic, so it can transfer more energy to the electrons.
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