A metal cathode whose work function is 3.3 eV is illuminated with 12 mW of light
ID: 1486975 • Letter: A
Question
A metal cathode whose work function is 3.3 eV is illuminated with 12 mW of light with a wavelength of 410 nm. Assume perfect efficiency of converting photons to photoelectrons. What current is measured in this experiment? If the power of the 410 nm light is increased to 67 mW, what current would be measured? If the wavelength of the 12 mW light were changed to 280 nm, what current would be measured? What is the maximum speed of an ejected electron if the power of the light is 12 mW and the wavelength is 280 nm? What is the stopping voltage when the power of the light is 12 mW and the wavelength is 280 nm? What is the cutoff wavelength of the cathode when the power of the light is 12 mW?Explanation / Answer
1)
Here , energy of 410 nm photon
E = 6.626 *10^-34* 3 *10^8/(410 * 10^-9)
E = 4.848 *10^-19 J
number of electrons ejected per second , N = 12 *10^-3/(4.848 *10^-19 )
N = 2.47 *10^16 electrons/sec
current flowing = N * e
current flowing = 2.47 *10^16* 1.602 *10^-19 C/s
current flowing = 3.965 *10^-3 C/s
current flowing = 3.965 mA
the current in the cathode is 3.965 mA
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2)---------------------------------------------------
Here , energy of 410 nm photon
E = 6.626 *10^-34* 3 *10^8/(410 * 10^-9)
E = 4.848 *10^-19 J
number of electrons ejected per second , N = 67 *10^-3/(4.848 *10^-19 )
N = 1.382 *10^17 electrons/sec
current flowing = N * e
current flowing = 1.382 *10^17* 1.602 *10^-19 C/s
current flowing = 22.14 *10^-3 C/s
current flowing = 22.14 mA
the current in the cathode is 22.14 mA
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3)--------------------------------------------------------
Here , energy of 410 nm photon
E = 6.626 *10^-34* 3 *10^8/(280 * 10^-9)
E = 7.099 *10^-19 J
number of electrons ejected per second , N = 12 *10^-3/(7.099 *10^-19 )
N = 1.6902 *10^16 electrons/sec
current flowing = N * e
current flowing = 1.6902 *10^16* 1.602 *10^-19 C/s
current flowing = 2.708 *10^-3 C/s
current flowing = 2.708 mA
the current in the cathode is 2.708 mA
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4)---------------------------------------------
let the maximum speed is v
energy of photon , E = 6.626 *10^-34* 3 *10^8/(280 * 10^-9)
E = 7.099 *10^-19 J
work function = 3.3 eV
0.5 * 9.11 *10^-31 * v^2 = 7.099 *10^-19 - 3.3 * 1.602*10^-19
v = 6.307 *10^5 m/s
the maximum speed of ejected electrons is 6.307 *10^5 m/s