Solve part d only When light of wavelength 307 nm is shined on a uranium metal s

Question

Solve part d only

When light of wavelength 307 nm is shined on a uranium metal surface, electrons are ejected from the metal with a velocity 5.4 times 105 m/s. This phenomenon is called the photoelectric effect. Calculate the frequency, in Hertz (Hz), of the light shined on the metal surface? 9.77 times 10^14 Hz What is the energy of a photon of this light in Joules (J)? 6.48·times 10^-19J What is the wavelength of the ejected electrons? 1.4 times 10-9 n in a separate experiment the kinetic energy of the emitted electrons was observed to be 2.45 times 10^-19 J. If the energy of the light shined on the uranium surface was 1.930 times 10^-18 J/photon, what is the binding energy (work function) of uranium? Give your answer in Joules (J).

Explanation / Answer

Wavelength of light = 307 nm = 307*10-9 m

Velocity of the ejected electron = 5.4*105 m/s

Mass of electron = 9.10*10-31 kg

12 a) Velocity of light = 3*108 m/s

Use the relation * = c where = frequency of the incident radiation; = wavelength of the radiation and c = velocity of radiation.

Plug in values:

*(307*10-9 m) = (3*108 m/s)

===> = (3*108)/(307*10-9) s-1 = 9.772*1014 s-1 = 9.772*1014 Hz (1 Hz = 1 s-1) (ans).

b) The energy of the photon of light is given by E = h* where h = Planck’s constant = 6.626*10-34 J.s.

Plug in values:

E = (6.626*10-34 J.s)*(9.772*1014 s-1) = 6.475*10-19 J (ans).

c) The maximum kinetic energy of the ejected electron is given by Kmax = ½*(9.10*10-31 kg)*(5.4*105 m/s)2 = 2.653*10-19 kg.m2 s-2 = 2.653*10-19 J.

We know that

Kmax = h*v – where = hc/0 is the work function of the metal and 0 is the wavelength of the ejected electron.

Plug in values:

2.653*10-19 J = 6.475*10-19 J – hc/0

===> hc/0 = 6.475*10-19 J – 2.653*10-19 J = 3.822*10-19 J

===> 0 = hc/(3.822*10-19 J) = (6.626*10-34 J.s)*(3*108 m/s)/(3.822*10-19 J) = 5.20*10-7 m = (5.20*10-7 m)*(109 nm/1 m) = 520 nm (ans).

d) Given KEmax = 2.45*10-19 J and energy of light shone = 1.930*10-18 J/photon.

Therefore,

2.45*10-19 J = 1.930*10-18 J – where = work function of the metal = binding energy of the metal.

===> = 1.930*10-18 J – 2.45*10-19 J = 1.685*10-18 J (ans).

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.