1) What is the energy of a 656 nm photon? A. 0.32 eV B. 1.89 eV C. 1.96 eV D. 3.

Question

1) What is the energy of a 656 nm photon? A. 0.32 eV B. 1.89 eV C. 1.96 eV D. 3.14 ev E. 5.1 e 2) What is a photon? A. an electron in an excited state B. a small packet of electromagnetic energy that has particle-like properties C. one form of a nucleon, one of the particles that makes up the nucleus D. an electron that has been made electrically neutral E. a proton that has been made electrically neutral 3) A metal surface is illuminated with blue light and electrons are ejected at a given rate each with a certain amount of energy. If the intensity of the blue light is increased, electrons are ejected A. at the same rate, but with more energy per electron. B. at the same rate, but with less energy per electron. C. at an increased rate with no change in energy per electron. D. at a reduced rate with no change in energy per electron. in cnergy per 4) Compared to a 2 eV photon, a 5 eV photon has A. lower frequency B. shorter wavelength C. longer wavelength D. higher speed E. lower speed 5) In order for a photon to eject an electron from a metal's surface in the photoelectric effect, the photon's A. frequency must be less than a certain minimum value. B. speed must be greater than a certain minimum value C. wavelength must be less than a certain minimum value. D. momentum must be zero. 6)A beam of red light and a beam of violet light each deliver the same power on a surface. For which beam is the number of photons hitting the surface per second the greatest? A. the red beam B. the violet beam C. The number of photons per second is the same for both beams. D. This cannot be answered without knowing just what the light intensity is. 7) In the Bohr model of the atom, energy is radiated when A. an electron falls from an outer energy level to an inner level B. an electron is stripped from an atom C. an ion is formed D. light shines on an atom E. an atom is in its ground state.

Explanation / Answer

1.

Energy

E=hc/lambda =(6.626*10-34)(3*108)/(656*10-9)

E=3.03*10-19 J

1 eV =1.6*10-19 J

E=(3.03*10-19)/(1.6*10-19)

E=1.89 eV

8.

KE =hc/lambda -W

KE =(6.626*10-34)(3*108)/(620*10-9)(1.6*10-19)-1.7

KE=0.3 eV

9.

Debroglie wavelength

lambda=h/p =(6.626*10-34)/(1.95*10-27)

lambda =340 nm

10.

Kinetic energy

KE=(1/2)mV2

1.6*10-19 =(1/2)*(9.11*10-31)*V2

V=5.92*105 m/s

momentum

p=mV =(9.11*10-31)(5.92*105) = 5.4*10-25

wavelength

lambda =h/p =6.626*10-34/(5.4*10-25) =1.2 nm

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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