The energy of a certain photon was calculated at 4.41 times 10^-19 J. What is th

Question

The energy of a certain photon was calculated at 4.41 times 10^-19 J. What is the wavelength (lambda) of this photon, in nm, and where is it approximately located in the electromagnetic spectrum? (c = 3 times 10^8 m/s) & (h = 6.626 times 10^-34 J- s) 625 nm (Visible Region; Orange) 863 nm (Infrared Region) 721 nm (Visible Region; Red) 451 nm (Visible Region; Blue) 222 nm (Ultraviolet Region) What is the wavelength (lambda), in nm, of a photon emitted during a transition from the n = 5 to the n = 2 state in the hydrogen atom? (R_H = 2.18 times 10^-18 J) 564 nm 235 nm 575 nm 434 nm 116 nm

Explanation / Answer

Answer for P10.

E = hcWavelength

Wavelength = hcE

= 6.62 X 10-34 X 3 X 108 m 4.41 X 10 -19

= 4.5 X 10-7 X 109 nm (1m = 109 nm)

= 450 nm (option d)

Answer for P11. 435 nm (option d)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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