StudentAidUnit 19: Dawn of Modern Physics — Numericals
12th Class Physics · Unit 19: Dawn of Modern Physics
19.1.The period of pendulum is measured to be 3.0 sec. in the inertial reference frame of the pendulum. What is its period measured by an observer moving at a speed of 0.95 c with respect to the pendulum?
Given
Proper time
t0 = 3.0 s
Velocity
v = 0.95c
Formula
Time dilation
t = frac{t0}{sqrt{1 - v2c2
Working
Substituting
t = frac{3.0}{sqrt{1 - (0.95)2c2
= frac{3.0}{sqrt{1 - 0.9025} = frac{3.0}{sqrt{0.0975} = 3.00.3122
Result
Time increased
t = 9.6 sec
19.2.A bar 1.0 m in length and located along x-axis moves with a speed of 0.75 c with respect to a stationary observer. What is the length of the bar as measured by the stationary observer?
Given
Proper length
l0 = 1.0 m
Velocity
v = 0.75c
Formula
Length contraction
l = l0 sqrt{1 - v2c2
Working
Substituting
l = 1.0sqrt{1 - (0.75)2c2
= 1.0sqrt{1 - 0.5625} = 1.0sqrt{0.4375} = 1.0 × 0.6614
Result
Length contraction
l = 0.66 m
19.3.Find the mass m of a moving object with speed 0.8 c.
Given
Speed of object
v = 0.8c
Formula
Relativistic mass
m = frac{m0}{sqrt{1 - v2c2
Working
Substituting
m = frac{m0}{sqrt{1 - (0.8)2c2
= frac{m0}{sqrt{1 - 0.64} = frac{m0}{sqrt{0.36} = m00.6
Result
Mass of object
m = 1.67 m0
19.4.Assuming you radiate as does a black body at your temperature about 37°C, at what wavelength do you emit the most energy?
Given
Temperature
T = 37°C + 273 = 310 K
Wien's constant
b = 2.9 × 10-3 mK
Formula
Wien's displacement law
lambdamax × T = Wien's constant
Working
Solving for wavelength
lambdamax = frac{2.9 × 10-3{310}
= 9.35 × 10-6 m
Result
Maximum wavelength
lambdamax = 9.35 μm
19.5.What is the energy of a photon in a beam of infrared radiation of wavelength 1240 nm?
Given
Wavelength
lambda = 1240 nm = 1240 × 10-9 m
Formula
Photon energy
E = hf
Frequency relation
f = clambda
Combined formula
E = hclambda
Working
Substituting
E = frac{6.63 × 10-34 × 3 × 108{1240 × 10-9
= frac{19.89 × 10-26{1.240 × 10-6 = 1.6 × 10-19 J
Result
Energy in eV
E = 1.0 eV
19.6.A sodium surface is illuminated with light of wavelength 300 nm. The work function of sodium metal is 2.46 eV. Find (a) Maximum energy of the ejected electron. (b) Determine the cutoff wavelength for sodium.
Given
Wavelength
lambda = 300 nm = 300 × 10-9 m
Work function
phi = 2.46 eV = 2.46 × 1.6 × 10-19 J
Formula
(a) Photon energy
E = hclambda
Working
Substituting values
E = frac{6.63 × 10-34 × 3 × 108{300 × 10-9
= 6.63 × 10-19 J = 4.14 eV
Formula
Maximum kinetic energy
K.Emax = hf - phi
(a) Calculating K.E.
= 4.14 - 2.46 = 1.68 eV
Result
(a) Maximum K.E.
K.Emax = 1.68 eV
Formula
(b) Cutoff condition
phi = hclambda0
Working
(b) Solving for cutoff
lambda0 = hcphi = frac{6.63 × 10-34 × 3 × 108{3.94 × 10-19
= 5.05 × 10-7 m = 505 × 10-9 = 505 nm
Result
(b) Cutoff wavelength
lambda0 = 505 nm
19.7.A 50 KeV photon is Compton scattered by a quasi-free electron. If the scattered photon comes off at 45° what is its wavelength?
Given
Initial energy
E = 50 KeV = 50 × 103 eV = 5 × 104 × 1.6 × 10-19 J
Scattering angle
θ = 45°
Formula
Initial wavelength
lambdai = hcE
Working
Calculating initial wavelength
lambdai = frac{6.63 × 10-34 × 3 × 108{5 × 104 × 1.6 × 10-19
= frac{19.89 × 10-26{8 × 10-15 = 2.486 × 10-11 m = 0.0248 nm
Formula
Compton shift
Δlambda = hme c(1 - cos θ)
Working
Calculating shift
Δlambda = frac{6.63 × 10-34{9.1 × 10-31 × 3 × 108(1 - cos 45°)
= frac{6.63 × 10-34-8+31{27.3}(1 - 0.707) = 0.2429 × 10-11(0.293) = 0.07 × 10-11 m = 0.0007 nm
Scattered wavelength
lambdas = lambdai + Δlambda = 0.0248 + 0.0007
Result
Scattered wavelength
lambdas = 0.0255 nm
19.8.A particle of mass 5 mg moves with speed 8.0 m/s. Calculate its de-Broglie wavelength.
Given
Mass of particle
m = 5 mg = 5 × 10-6 kg
Speed
v = 8 m/s
Formula
de Broglie wavelength
lambda = hmv
Working
Substituting
lambda = frac{6.63 × 10-34{5 × 10-6 × 8.0}
= frac{6.63 × 10-34{40 × 10-6 = 6.6340 × 10-28 = 1.66 × 10-29 m
Result
Wavelength
lambda = 1.66 × 10-29 m
19.9.An electron is accelerated through a potential difference of 50 V. Calculate its de-Broglie wavelength.
Given
Potential difference
V = 50 V
Formula
de Broglie wavelength for accelerated electron
lambda = frac{h}{sqrt{2mVe}
Working
Substituting
lambda = frac{6.63 × 10-34{sqrt{2 × 9.1 × 10-31 × 50 × 1.6 × 10-19
= frac{6.63 × 10-34{sqrt{1.456 × 10-47 = frac{6.63 × 10-34{3.817 × 10-24
Result
Wavelength
lambda = 1.74 × 10-10 m
19.10.The life time of an electron in an excited state is about 10^-8 s. What is its uncertainty in energy during this time?
Given
Lifetime
Δt = 10-8 s
Formula
Uncertainty principle
ΔE · Δt ≈ h
Working
Solving for energy uncertainty
ΔE = hΔt = frac{1.05 × 10-34{10-8
Result
Energy uncertainty
ΔE = 1.05 × 10-26 J
19.11.An electron is to be confined to a box of size of the nucleus (1 × 10^-14 m). What would the speed of electron be if it were so confined?
Given
Confinement size
Δx = 1 × 10-14 m
Formula
Position-momentum uncertainty
ΔP ≈ hΔx
Velocity from momentum
m ΔV = hΔx
Working
Solving for velocity
ΔV = hm Δx = frac{1.05 × 10-34{9.1 × 10-31 × 1 × 10-14
= 1.059.1 × 10-34+31+14 = 0.115 × 1011 = 1.15 × 1010 m/s
Result
Speed uncertainty
ΔV = 1.15 × 1010 m/s