Kinetic energy of the emmited electrons

[WAHEEDA]

(1) The work function of a particular metal is 2.00 eV.
(a) If the metal is illuminated with monochromatic light of wavelength 550 nm, what will be the maximum kinetic energy of the emmited electrons and,
(b) their maximum speed?
(c) What is the required stopping potential?
(d) What are the threshold frequency and corresponding wavelength for
the metal?

(2) A potential difference of 40kV is applied across an X-rays radiated?please I need help..and need the answer by this coming monday,29th november 2004!

 

[chroot]

The work function is the amount of energy required to remove an electron from a metal. The amount of energy contained in a photon of light of frequency v is E = hv. Some of that energy is used to liberate the electron from the metal; the rest ends up as kinetic energy of the electron. You should be able to calculate the speed from the kinetic energy.

- Warren

 

[wais]

(a) The maximum kinetic energy of the emitted electrons can be calculated using the equation: KEmax = h*c/λ - Φ, where h is Planck's constant, c is the speed of light, λ is the wavelength of the light, and Φ is the work function of the metal. Plugging in the given values, we get KEmax = (6.626*10^-34 J*s * 3.00*10^8 m/s)/(550*10^-9 m) - 2.00 eV. Converting the electron volts to joules, we get KEmax = 3.60*10^-19 J.

(b) The maximum speed of the emitted electrons can be calculated using the equation: v = √(2KEmax/m), where m is the mass of an electron. Plugging in the values, we get v = √(2*3.60*10^-19 J/9.11*10^-31 kg) = 6.61*10^5 m/s.

(c) The required stopping potential can be calculated using the equation: Vstop = KEmax/e, where e is the charge of an electron. Plugging in the values, we get Vstop = (3.60*10^-19 J)/(1.602*10^-19 C) = 2.25 V.

(d) The threshold frequency can be calculated using the equation: f = Φ/h, where f is the threshold frequency. Plugging in the values, we get f = (2.00 eV)/(6.626*10^-34 J*s) = 3.02*10^15 Hz. The corresponding wavelength can be calculated using the equation: λ = c/f, where c is the speed of light. Plugging in the values, we get λ = (3.00*10^8 m/s)/(3.02*10^15 Hz) = 9.93*10^-8 m or 99.3 nm.

(2) The energy of the X-rays can be calculated using the equation: E = qV, where E is the energy, q is the charge, and V is the potential difference. Plugging in the values, we get E = (1.602*10^-19 C)*(40*10^3 V) = 6.41*10^-15 J.