Amodern compact fluorescent lamp contains 1.4 mg of mercury. if each mercury atom in the lamp were to emit asingle photon of wavelength 254 nm, how many joules of energy would be emitted?

Energy emitted : 3,0714 J

The photoelectric effect is an electron coming out of a metal because of electromagnetic radiation

One type of electromagnetic radiation is light

Electrons can come out of metal because they absorb electromagnetic energy radiated on metals. There is also kinetic energy released from metal, which is according to the equation:

fo = the threshold frequency of electromagnetic waves

Radiation energy is absorbed by photons

The energy in one photon can be formulated as

Where

E = energy of light, J

h = Planck's constant (6,626.10⁻³⁴ Js)

f = Frequency of electromagnetic waves, Hz

f = c / λ

c = speed of light

= 3.10⁸

λ = wavelength

We determine the number of mercury atoms (Hg) in 1.4 mg

atomic mass Hg = 200

mass Hg = 1.4 mg = 1.4 .10⁻³g

mol Hg = mass: atomic mass

mol Hg = 1.4.10⁻³ g: 200

mol Hg = 7.10⁻⁶

Number of atoms: mole x Avogadro number

(1 mole = 6.02.10²³ particles)

number of atoms: 7.10⁻⁶ x 6.02.10²³

number of atoms: 4.214 x10¹⁸

Frequency of mercury rays:

f = c: λ

f = 3.10⁸: 254 .10⁻⁹

f = 1.1.10¹⁵

The radiation emitted by 1 mercury atom

E = h. f

E = 6,626.10⁻³⁴. 1.1.10¹⁵

E = 7.2886.10⁻¹⁹ J

So that the total energy

E total = 4,214.10¹⁸ .7.2886.10⁻¹⁹

E total  = 3,0714 J

Electromagnetic radiation with wavelength of 745 nm  

link  

The energy of a photon  

link  

The equation E = hf  

link  

the approximate energy of a photon  

link  

the correct answer is boyle