Problems on Light The minimal frequency for photoelectric emission in calcium is

Question

Problems on Light

The minimal frequency for photoelectric emission in calcium is 7.7 times 10 14 Hz. Find the maximum KE of the electrons emitted when light of frequency 12.0 times 10 15 Hz is directed on a calcium surface. 4 times l0 -19 J of energy is required to remove and electron from the surface of a particular metal. What is the frequency of light that will just dislodge electrons from the surface? What is the maximum energy of electrons emitted through the action of light of wavelength 2 times 10 -7 m? Electrons are accelerated through potential differences of approximately 10,000 V in television picture tubes. Find the maximum frequency of the x-rays that are produced when these electrons strike the screen of the tube. (1.6 times 10 -19 J=1 eV)

Explanation / Answer

6. work function = 120-7.7 = 112.3*10^14 Hz

KE max = hf = 112.3*10^14 * 6.626*10^-34  

KE max = 74.4 *10^-19 J or 46.5 eV

7. E = hf

frequency f = E/h = 4*10^-19/6.626*10^-34 =6.03 *10^14 Hz

E = hc/L

E = 6.626*10^-34 * 3*10^8/(2*10^-7)

E = 9.939 *10^-19 J or 6.211 eV

8. KE = 0.5 mv^2 = eV = 1.6 *10^-19 *10000 = 1.6 *10^-15 J

noW E = hf

fmax = 1.6 *10^-15/6.626*10^-34

fmax = 2.414*10^18 Hz

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