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To answer this question, we will use the following formula:
hf = phi + Emax 

where 
h = plancks constant = 6.626x10^-34 J s 
f = frenquency of incident light = 1.6×10^15 Hz
phi = work function of the cesium = 2.14 eV = 3.428x10^-19 J
Emax = maximum kinetic energy of the emitted electrons. 

Substitute with the givens in the above equation to calculate Emax as follows:
6.626x10^-34 x 1.6×10^15 = 3.428x10^-19 + Emax
Emax = 7.1736 x 10^-19 J
BladeRunner212

7.248 x 10⁻¹⁹ J

Further explanation

Einstein conveyed his idea of light acting as particles. The explanation is included in the photoelectric effect.

  • The work function of the metal, denoted by [tex]\boxed{ \ \phi \ }[/tex], is a measure of the minimum energy required to escape an electron from the surface of solid.
  • The UV Light Energy arrives in energy packages called photons (or photoelectrons).

The energy carried by a photon is given by [tex]\boxed{\boxed{ \ E = hf \ }}[/tex]

E = energy in joules

h = Planck's constant [tex]\boxed{ \ 6.63 \times 10^{-34} \ Js^{-1} \ }[/tex]

f = frequency of light (in Hz)

  • The threshold frequency f₀ is the frequency at which no photoelectrons are emitted.
  • Above f₀, the maximum kinetic energy of these metal electrons depends on the frequency of the incident light onto a metal surface.
  • The number of electrons emitted depends on the intensity of the light and does not depend on the frequency.

Above the threshold frequency, incoming energy of photons equal the work function plus kinetic energy (KE).

[tex]\boxed{\boxed{ \ hf = \phi + KE \ }}[/tex]

Given that:

  • Frequency of UV rays [tex]\boxed{ \ f = 1.6 \times 10^{15} \ Hz \ }[/tex]
  • The work function of cesium [tex]\boxed{ \ \phi = 2.1 \ eV \ } \rightarrow \boxed{ \ \phi = 2.1 \times 1.6 \times 10^{-19} = 3.36 \times 10^{-19} \ J \ }[/tex]

Question:

The kinetic energy of the emitted electrons of cesium.

The process:

[tex]\boxed{ \ KE = hf - \phi \ }}[/tex]

[tex]\boxed{ \ KE = (6.63 \times 10^{-34})(1.6 \times 10^{15}) - 3.36 \times 10^{-19} \ }}[/tex]

Thus the kinetic energy of electron is [tex]\boxed{\boxed{ \ 7.248 \times 10^{-19} \ J \ }}[/tex]

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Notes:

To calculate the maximum speed electrons emitted from a cesium surface, use the formula:

[tex]\boxed{ \ v = \sqrt{\frac{2 \ KE}{m}} \ }[/tex]

[tex]\boxed{ \ with \  m \ = 9.1 \times 10^{-31} \ }[/tex]

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Keywords: the kinetic energy, the emitted electrons, cesium, UV rays, frequency, Einstein, photoelectric, photoelectrons, photons, Planck, the work function, threshold

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