IV. Classical Catastrophes & The Beginnings of Quantum Mechanics In these proble

Question

IV. Classical Catastrophes & The Beginnings of Quantum Mechanics In these problems, it is assumed that you know the Wien displacement law, which I do not mention in my notes. This is derived from the Planck blackbody distribution, and states that the radiation emitted from a blackbody peaks at a wavelength given by max-b/T, where T is measured in Kelvin, and b-2.8978 × 10-3 m. K. 8. The fireball in a thermonuclear explosion can reach temperatures of approximately 10"K What value of max does this correspond to? In what region of the spectrum is this wavelength found? (The book refers to a figure in the chapter, Fig. 1.11. Just Google "electromagnetic spectrum" and you'll find what you need here.) 9. A household light bulb is a blackbody radiator. Many light bulbs use tungsten filaments that are heated by an electric current. What temperature is needed so that max= 550 nm? 10. My green laser pointer (a-532 nm) has a power output of 1 milliWatt (mW). While pointing at the board for one second, how many photons were emitted?

Explanation / Answer

Q8.

a) Find lamda max given:

T = 10^7 K

lamda max = ?

the formula given

lamda max = b/T

b = 2.8978*10^-3

then, substitute data

lamda max = b/T

lamda max = (2.8978*10^-3)/(10^7) = 2.8978e*10^-10 m

lamda max to nanometer (10^9) = (2.8978e*10^-10)(10^9) = 0.7877037 nanometers

b) region of electromagnetic spectrum

since 0.01 to 10 nanometers is the range for x-rays, this must be in the X-Rays range since

0.01 < 0.78 < 10 nanometer

Q9.

Apply formula

lamda max = b/T

solve for T

T = b/ lamda max

T = ((2.8978*10^-3))/(550*10^-9) K

T = 5268.7272 K

we need a temperature about 5268.7272K

For further questions, pleas consider posting multiple questions in multiple set of Q&A.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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