Homework 1 Key

Last updated
Save as PDF

Page ID: 109898

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

Q1.1

Rigel:

λ_max T=2.9x10^-3mK

λ_max=145 nm=1.45x10^-7 m

T= (2.9x10^-3mK)/(1.45x10^-7m)= 2.00x10⁴ K

Betelgeuse:

T= (2.9x10^-3mK)/(7.00x10^-7m)= 4.14x10³ K

The surface temperature of Betelgeuse is cooler by an order of magnitude.

Q1.2

KE = hν - ϕ

0 = hν-5.22eV

5.22eV=(ν)x(4.14x10^-15 eV·s)

ν= 1.26x10¹⁵ Hz

λ= c/ν= (3x10⁸ m/s)/(1.26x10¹⁵ Hz)= 2.38x10^-7 m= 238 nm

This falls in the UV region of the electromagnetic spectrum.

Q1.3

1) The intensity of light is 40 W, so the energy of the light that shines from the LED for a 1 second period is: 40 W x 1 s = 40 Joules.

ν= c/λ= (3x10⁸ m/s)/(5.64x10^-7 m)= 5.32x10¹⁴ Hz

E_photon= hν= (6.63×10^-34 Js)x(5.32x10¹⁴ Hz) = 3.53x10^-19 J

The number of photons emitted per second is: (40 J)/(3.53x10^-19 J)= 1.1x10²⁰

2) The intensity of the radiation is 1200 W, so the energy emitted by the microwave in a 1 second period is: 1200 W x 1 s = 1200 Joules.

ν= c/λ= (3x10⁸ m/s)/(1.1x10^-2 m)= 2.7x10¹⁰ Hz

E_photon= hν= (6.63×10^-34 Js)x(2.7x10¹⁰ Hz) = 1.8x10^-23 J

The number of photons emitted per second is: (1200 J)/(1.8x10^-23 J)= 6.7x10²⁵

Q1.4

a) KE = hν - ϕ

hν = KE + ϕ = 200eV + 5.0eV = 205eV = hc/λ

λ = (4.14x10^-15 eV s)x(3x10⁸ m/s)/(205eV) = 6.06x10^-9 m = 6.06 nm

b) (6.0 W)x(3600 s) = 21,600 J

E_photon= KE + ϕ =(205eV)x(1.6x10^-19J/eV)= 3.3x10^-17J

(21,600 J total energy)/(3.3x10^-17J per electron)= 6.5x10²⁰ electrons can be ejected in total.

c) (1.00x10^-10J total energy)/(3.3x10^-17J per electron)= 3.0x10⁶ electrons can be ejected in total.

Q1.5

a) KE = 0 = hν - ϕ

hν = hc/λ = ϕ

λ = hc/ϕ = (4.14x10^-15 eV s)x(3x10⁸ m/s)/(4.59eV)= 2.7x10^-7m= 270 nm

b) 500 nm is not enough to eject an electron, minimum required energy is 270 nm. \(KE_{impure} = KE_{pure} = 0\). If the wavelength was shorter, the maximum KE for impure C60 will be greater than that for pure C60.

Q1.6

a) 3.00 mW= 3.00 mJ/s

E_photon= hc/λ =(6.63×10^-34 Js)x(3x10⁸ m/s)/(400x10^-9m)= 4.97x10^-19J

The number of photons incident per second = the number of electrons ejected per second = (3.00x10^-3 J/s)/(4.97x10^-19J) = 6.04x10¹⁵ electrons ejected per second.

b) KE = hν - ϕ = 4.97x10^-19J - (2.3eV)x(1.6x10^-19J/eV)= 1.29x10^-19J

Total power = KE x number of electrons ejected/second = 7.74x10^-4 W= 0.774 mW

Q1.7

1/λ=R_H(1/2² - 1/3²)= R_H(1/4 - 1/9)= 5R_H/36

λ= 36/5R_H= 36/(5x1.1x10^-7 m)=7.9x10^-7m= 790 nm (655 nm? Julia)

Q1.8

Calculate the energy of each photon:

E_photon= hν = hc/λ =(6.63×10^-34 Js)x(3x10⁸ m/s)/(1.12x10⁴x10^-9m)= 1.77x10^-20J

Calculate the energy needed to heat 266 g of water by 16 degrees Celsius:

ΔE_water = mCΔt = (266g)×(4.184 J/g°C)x(16°C) = 1.78x10⁴ J

The number of photons needed is: ΔE_water/ E_photon = 1.0x10²⁴

Q1.9

C_fusionm_ice = (number of photons) × E_photon = (number of photons) × hc/λ

m_ice = (number of photons) x (hc/λ) x (1/C_fusion) = (4.0x10²³)x(6.63×10^-34 Js)x(3x10⁸ m/s)/(150x10^-9m)x(333.55x10³J/kg) = 1.59kg =1590 g

Mass of each water molecule: (18 g/mol)x(1 mol/N_A) = 18 g/6.022x10²³ = 2.99x10^-23 g

Therefore the number of water molecules that melted is: m_ice / m_{water molecules} = 1590 g / 2.99x10^-23 g = 5.32x10²⁵

Q1.10

The Planck distribution is represented by the solid black line overlaid with red dashes (1) that converges on both sides. The curve for the toast cooked at the ideal temperature would be shifted right (cooler) on a plot versus wavelength; the inverse is true for a plot versus frequency. The line showing the Wien displacement law is represented by 3 (the leftmost dashed lines that diverge), and the Rayleigh-Jeans' Law is represented by 2.

λ_max = b / T= 2.9x10^-3mK / (154+273)K= 6.79x10^-6 m= 6790 nm

Since the highest intensity radiation falls in the IR range of the electromagnetic spectrum (not the visible), the toast itself is not expected to glow. Luckily, this hypothesis is supported by experimental observations.

(Photo credit: Dr. Koski; powerpoint from 9/29 lecture)