Question

Cesium is often used in “electric eyes” for self-opening doors in an application of the photoelectric effect. The amount of energy required to ionize (remove an electron from) a cesium atom is 3.89 electron volts (). Show by calculation whether a beam of yellow light with wavelength 5830 Å would ionize a cesium atom.

Energy of the light beam = _____eV

Would a beam of yellow light with wavelength 5830 Å ionize a cesium atom? _____(No/Yes)

Answer 1

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Answer 2

To calculate the energy of a photon of yellow light with a wavelength of 5830 Å, we can use the equation:

E = hc/λ

where h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (2.998 x 10^8 m/s), and λ is the wavelength in meters.

First, we need to convert the wavelength from angstroms to meters:

5830 Å = 5830 x 10^-10 m

Now we can plug in the values and solve for E:

E = (6.626 x 10^-34 J s)(2.998 x 10^8 m/s)/(5830 x 10^-10 m) E = 3.22 x 10^-19 J

To convert joules to electron volts, we can use the conversion factor:

1 eV = 1.602 x 10^-19 J

E = (3.22 x 10^-19 J)/(1.602 x 10^-19 J/eV) E = 2.01 eV

The energy of the yellow light beam is 2.01 eV. Since this is less than the energy required to ionize a cesium atom (3.89 eV), the yellow light beam would not ionize a cesium atom. Therefore, the answer is No.