Titanium metal requires a photon with a minimum energy of 6.94×1019J to emit ele

Question

Titanium metal requires a photon with a minimum energy of 6.94×1019J to emit electrons.

Part A

What is the minimum frequency of light necessary to emit electrons from titanium via the photoelectric effect?

Express your answer using three significant figures.

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Part B

What is the wavelength of this light?

Express your answer using three significant figures.

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Part C

This question will be shown after you complete previous question(s).

Part D

If titanium is irradiated with light of 233 nm, what is the maximum possible kinetic energy of the emitted electrons?

Express your answer using three significant figures.

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Titanium metal requires a photon with a minimum energy of 6.94×1019J to emit electrons.

Part A

What is the minimum frequency of light necessary to emit electrons from titanium via the photoelectric effect?

Express your answer using three significant figures.

=   s1  

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Part B

What is the wavelength of this light?

Express your answer using three significant figures.

=   nm  

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Part C

This question will be shown after you complete previous question(s).

Part D

If titanium is irradiated with light of 233 nm, what is the maximum possible kinetic energy of the emitted electrons?

Express your answer using three significant figures.

E =   J  

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Explanation / Answer

Part A :

minimum energy to emit electrons E = 6.94×1019 J

E = hv

h = 6.625*10^-34 j.s

v= frequency = ?

(6.94*10^(19)) = (6.625*10^(-34))*v

v = 1.05*10^15 s-1

Part B: wavelength of this light,

C = vL

c= light velocity = 3*10^8 m/s

v = frequency = 1.05*10^15 s-1

L = wavelength = ?

3*10^8 = (1.05*10^15)*L

L = 2.86*10^(-7) m = 286 nm

Part D

from photoelectric effect

hv = hv0 + K.E

K.E = (6.625*10^(-34)*(3*10^8/(233*10^(-9)))-(6.625*10^(-34)*1.05*10^15)

   K.E = 1.57*10^-19 joule.

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