Engineering Physics B.E. First Semester / 20 Practice Questions PDF

## Engineering Physics B.E. First Semester / 20 ### Theoretical Calculation of λ By using de Broglie hypothesis, the wavelength of an electron can be calculated as: $λ_{theoretical} = \frac{12.27}{√V}= \frac{12.27}{√54} =1.67 Å$ The agreement between theoretical and experimental values obtained from...

## Engineering Physics B.E. First Semester / 20 ### Theoretical Calculation of λ By using de Broglie hypothesis, the wavelength of an electron can be calculated as: $λ_{theoretical} = \frac{12.27}{√V}= \frac{12.27}{√54} =1.67 Å$ The agreement between theoretical and experimental values obtained from different considerations is excellent and constitutes a convincing proof for the wave characteristics of electron and hence the de Broglie hypothesis. ### 1.6 List of Formulae * Compton shift in terms of modified and unmodified component: $Δλ = λ' - λ$ * Compton shift in terms scattering angle: $Δλ = \frac{h}{m_0c} (1- cosφ)$ * Modified wavelength: $λ' = λ + \frac{h}{m_0c} (1-cosφ) $ * Unmodified wavelength: $λ = λ' - \frac{h}{m_0c} (1- cosφ) $ * Energy of recoiled electron: $E = \frac{hcΔλ}{λλ'}$ * Velocity of recoil electron: $v = √\frac{2E}{m}$ * de Broglie wavelength in terms of velocity/ momentum: $λ = \frac{h}{mv} = \frac{h}{p}$ * de Broglie wavelength in terms of kinetic energy: $λ = \frac{h}{√2mE}$ * de Broglie wavelength in terms of accelerating potential: $λ = \frac{h}{√2meV}$ * de-Broglie wavelength of an electron: $λ = \frac{12.27}{√V} * 10^{-10}$ * Diffraction equation (Bragg's law): $2d sin θ= ηλ $ ## Engineering Physics B.B. First Semester /21 ### 1.7 Solved Numerical 1. X-rays of wavelength 0.5 A° are scattered by free electrons in a block of carbon through 90°. Find the kinetic energy and velocity of recoiled electrons. Solution: λ = 0.5 A°, φ = 90°, K.E. = ?, v =? $λ' = λ + \frac{h}{m_0c}(1-cosφ) $ $λ' = 0.5 + 0.02427 (1 - cos 90) = 0.52427 A°$ $Δλ = λ' - λ = 0.02427 A°$ $E = \frac{hcΔλ}{λλ'} $ $E = \frac{6.626×10^{-34}×3×10^8×0.02427×10^{-10}}{0.5×10^{-10} ×0.52427×10^{-10}} = 1.84×10^{-16} J$ $V = √\frac{2×K.E}{m} = √\frac {2×1.84×10^{-16}}{9.1×10^{-31}} = 2×10^7 m/sec.$ 2. Calculate the de Broglie wavelength of the orbital electron of hydrogen atom given that its energy is 13.6 eV. Solution: E = 13.6 eV, λ = ? $λ = \frac{h}{√2mE} = \frac{6.626×10^{-34}}{√2 × 9.1×10^{-31}×13.6×1.602×10^{-19}}=3.327×10^{-10}m = 3.327 A°$ 3. Calculate the de-Broglie wavelength associated with an alpha particle accelerated by a potential difference of 100V. Solution: V =100V, m = 6.68x10^-27kg, q = 2e= 2x1.6x10^-19C, λ = ? $λ = \frac{h}{√2mqV} = \frac{6.632×10^{-34}}{√2×6.68×10^{-27}×2×1.602×10^{-19}×100} = 1.014×10^{-13}m.$ 4. A bullet of mass 45 gms and an electron both travel with velocity of 1000m/s. What wavelength can be associated with them? What Inference can be drawn from this result? Solution: m=45 gms = 45x10^-3 Kg v = 1000m/s $λ_{bullet} = ? $, $λ_{electron} = ?$ $λ_{bullet} = \frac{h}{mv} = \frac{6.632×10^{-34}}{45×10^{-3}×1000}= 1.474×10^{-35} m = 1.474×10^{-25} Å$ $λ_{electron} = \frac{h}{mv} = \frac{6.632×10^{-34}}{9.1×10^{-31}×1000} = 7.2799×10^{-7} m = 7279.9×10^{-10} m = 7279.9 Å$ Inference: Since the wavelength of matter wave associated with the bullet is of the order of 10^-35m, the wave nature cannot be revealed. But, the wavelength of the matter wave associated with electron is large almost 10 times more than the size of the electron (10^-15m), hence the wave nature of electron is revealed in diffraction experiment. 5. A photon of energy 1MeV is scattered through 90º by a free electron. Calculate the change in energy of photon and electron after the interaction. Solution: E = 1 MeV = 1x10^6 eV = 1.602 X 10^-13 J φ = 90° E_{recoil}? $λ = \frac{hc}{E} = \frac{6.63×10^{-34}×3×10^8}{1.602×10^{-13}} = 12.41×10^{-13} m$ $λ' - λ = \frac{h}{m_0c}(1-cosφ)$ $λ' = λ + \frac{6.63×10^{-34}}{9.1×10^{-31}×3×10^8}(1-cos90°) $ $λ' = 3.66×10^{-12} m $ $E' = \frac{hc}{λ'} = \frac{6.63×10^{-34}×3×10^8}{3.66×10^{-12}} = 5.442×10^{-14} J$ Energy of recoil electron = E_{recoil} = E - E' = 1.057×10^-13 J 6. What would be the de-Broglie wavelength associated with (i) 2000 kg car having a constant speed of 25 m/s. (ii)80 kg scooter having a speed of 10m/s. Give your conclusion. Solution: For car: m=2000 Kg V=25 m/s For Scooter: m=80 kg V= 10m/s $λ_{car} = \frac{h}{mv} = \frac {6.632×10^{-34}}{2000×25}= 1.324×10^{-38} m$ $λ_{scooter} = \frac{h}{mv} =\frac {6.632×10^{-34}}{80×10}= 8.295×10^{-37} m$ Conclusion: Wavelength associated with car is less due to its larger mass. ## Engineering Physics B.E. First Semester /22 7. An x-ray photon of wavelength 0.5A is scattered through an angle 60º by a loosely bound electron. Calculate recoil energy and velocity of an electron. Solution: λ=0.5A, θ=60° $ \frac{h}{m_0c} = 0.024264 A°$ The energy of the recoil electrons should be equal to the difference in energies of the incident and scattered photons. Thus $E-E' = K.E.= hc(\frac{1}{λ} - \frac{1}{λ'}) = \frac{h}{m_0c} (\frac{1}{λ} - \frac{1}{λ'})$ $λ' = λ + \frac{h}{m_0c}(1 - cos θ)$ $λ = λ + (1 - cos θ) = 0.5 + 0.2426(1 - cos 60°) = 0.6213 A = 0.512 A $ $K.E = hc(\frac{1}{λ} - \frac{1}{λ'}) = 12400 (\frac{1}{λ} - \frac{1}{λ'}) = 12400 (1 - \frac{λ}{λ'})×1.602×10^{-19} J$ $K.E = 12400 (\frac{1}{0.5} - \frac{1}{0.6213} )×1.602×10^{-19} = 1.84×10^{-16} J$ $V = √\frac{2×K.E}{m} = √\frac{2×1.84×10^{-16}}{9.1×10^{-31}} = 2×10^7 m/s$ 8. An electron and a particle of mass 10 gm both are moving with velocity 300m/s. Calculate the wavelengths associated with them and interpret the results. Solution: m = 10gm, v = 300 m/s, λ = ?, λ' = ? $λ_{e} = \frac{h}{mv} = \frac{6.6260 ×10^{-34}}{10 × 10^{-3} × 300} = 2.2086 × 10^{-34} m = 2.2086 × 10^{-24} A°.$ $λ_{p} = \frac{h}{mv} = \frac{6.6260 ×10^{-34}}{9.1×10^{-31} × 300} = 2.4244 × 10^{-6} m$ Interpretation: Since the wavelength of the matter wave associated with the particle is very small of the order of 10^-24 A and therefore the wave nature of the particle is not revealed through the diffraction experiment. But the wavelength of matter wave associated with electron is large almost 10^5 times more than the size of the electron (10^-15m), hence the wave nature of electron is revealed in diffraction experiment. ## Engineering Physics B.E. First Semester / 23 9. Incident radiation of wavelength 1.087A is scattered from a scatterer at an angle of 30°. Calculate the wavelength of scattered photon and kinetic energy of recoil electron. Solution: λ =1.0872 A°, φ = 30°, λ' = ?, Κ.Ε. = ? $λ' = λ + \frac{h}{m_0c} (1 - cos φ) = 1.0872 + 0.02427(1 - cos 30°) = 1.09045 A.$ $E = \frac{hcΔλ}{λλ'} = \frac{6.626×10^{-34}×3×10^8×0.02427×10^{-10}}{1.0872×10^{-10} ×1.09045×10^{-10}} = 4.069×10^{-17} J.$ 10. Calculate the wavelength associated with a stone of mass 50gms moving with speed of 50m/s and an electron with kinetic energy of 100eV. Solution: m-50gms = 0.5 kg v = 50m/s K.E = 100eV $λ_{stone}= ?$, $λ_{e} = ?$ $λ_{stone} = \frac{h}{mv} = \frac{6.626×10^{-34}}{0.5×50}= 2.65×10^{-35} m$ $λ_{e}= \frac{h}{√2mE} = \frac {6.626×10^{-34}}{√2×9.1×10^{-31} ×1.602×10^{-19} ×100} = 1.224 A°$ 11. X-ray photon of wavelength 0.3A is scattered through an angle 40º by a loosely bound electron, find the energy of scattered photon. Solution: λ = 0.3A° φ = 40° E' = ? $λ' = λ + \frac{h}{m_0c} (1 - cos φ)$ $λ' = 0.3 + 0.02427(1 - cos 40) = 0.3056 Å = 0.3056 × 10^{-10} m $ $E' = \frac{hc}{λ'} = \frac{6.64×10^{-34} ×3×10^8}{0.3056×10^{-10}} = 6.5183 × 10^{-15} J$ ## Engineering Physics B.E. First Semester / 24 12. A bullet of mass 50 grams and an electron both travel with a velocity of 1000m/s. What wavelength can be associated with them? Solution: m₁ = 50 gms = 50 x 10^-3 kg v= 1000m/s m = 9.11 x10^-31 kg λ with bullet and electron = ? For bullet: $λ_b = \frac{h}{mv} = \frac{6.64×10^{-34}}{50 × 10^{-3} × 1000} = 0.1328 × 10^{-35} m$ For electron: $λ_e = \frac{h}{mv} = \frac{6.64×10^{-34}}{9.1 × 10^{-31} × 1000} = 7.288 × 10^{-7} m$ 13. X-ray photon of wavelength 1A are scattered from a carbon block in a direction 90°. Calculate the observed Compton shift. How much kinetic energy is imparted to the recoil electron? Solution: λ = 1A°, φ = 90°, Δλ = ?, Κ.Ε. = ? $Δλ = λ' - λ = \frac{h}{m_0c} (1 - cos φ) $ $λ' = λ + \frac{h}{m_0c} (1 - cos φ) $ $λ' = 1 + 0.02427(1 - cos90) = 1.02427 A°$ $E = \frac{hcΔλ}{λλ'} = \frac{6.626×10^{-34}×3×10^8×1.02427 × 10^{-10}}{1 × 10^{-10} × 1.02427 × 10^{-10}} = 1.9878 × 10^{-15} J$ 14. A bullet of mass 45 gm and an electron both travel with a velocity of 1200m/s. What wavelength can be associated with them? Why is the wave nature of bullet not revealed through diffraction effect? Solution: m = 45 gms = 45x10^-3 Kg v = 1200m/s $λ_b = ?$, $λ_e = ?$ $λ_{bullet} = \frac{h}{mv} = \frac{6.632×10^{-34}}{45×10^{-3}×1200} = 1.228 * 10^{-35} m = 1.228 * 10^{-25} A°$ $λ_{electron} = \frac{h}{mv} = \frac{6.632×10^{-34}}{9.1×10^{-31}×1200} = 6.07326×10^{-7} m = 6073.26×10^{-10}m= 6073.2A$ Inference: Since the wavelength of matter wave associated with the bullet is of the order of 10^-35m, therefore the wave nature cannot be revealed in diffraction experiment. ## Engineering Physics B.E. First Semester / 25 15. In Compton effect, a photon is scattered by a free electron at rest through an angle of 90°. What is the energy of scattered photon if the wavelength of incident photon is 12nm. Solution: $Δλ = \frac{h}{m_0c} (1 - cosφ) = 0.02426 (1-cos 90) $ $ = 0.02426 A = 0.2426 nm$ $λ' = λ + Δλ = 12.2426 nm $ $E' = \frac{hc}{λ'}= \frac{6.63 × 10^{-34} × 3 × 10^8}{12.2426 × 10^{-9}} = 1.62 × 10^{-17} J = 101.62 eV$. 16. Find the de-Broglie wavelength of: i) An electron accelerated through a potential difference of 144V. ii) 10gm object moving with a speed of 10m/s Solution: m=10gms = 0.1kg V=144V V=10m/s $λ_{object} = ?$, $λ_{e} = ?$ $λ_{object} = \frac{h}{mv} = \frac {6.626×10^{-34}}{10×10×10} = 6.63×10^{-33} m$ $λ_{e} = \frac{h}{√2meV} = \frac{6.626×10^{-34}}{√2×9.1×10^{-31} ×1.602×10^{-19} ×144} = 1.0225 A°$ 17. An X-ray photon of wavelength 0.3 Aº is scattered through an angle of 45º by a loosely bound electron. Find the wavelength of the scattered photon. Solution: λ = 0.3 A° φ = 45° λ' = ? $λ' = λ + \frac{h}{m_0c} (1 - cos φ)$ $λ' = 0.3 + \frac{6.626×10^{-34}}{9.1×10^{-31} *3 * 10^8}(1 - cos 45) = 0.3071 A° $ 18. A beam of X-rays are scattered by loosely bounded electrons at 45° from the direction of the beam. The wave length of the scattered X-rays is 0.22 A.What is the wavelength of incident X-rays. Solution: λ' = 0.22 A° λ = ? φ = 45° $λ = λ' - \frac{h}{m_0c} (1 - cos φ) $ $λ = 0.22 - \frac{6.626×10^{-34}}{ 9.1×10^{-31} * 3 * 10^8} (1 - cos 45) = 0.212 A $ ## Engineering Physics B.E. First Semester / 26 19. Calculate the wavelength of scattered radiation and velocity of recoil electron when wavelength of incident radiation is 1.0872A and angle of scattering is 30º. Solution: λ = 1.0872 A° φ = 30° λ' = ? v = ? $λ' = λ + \frac{h}{m_0c} (1 - cos φ) = 1.0872 + 0.02427(1 - cos 30) = 1.09045 A° $ $E = \frac{hcΔλ}{λλ'} = \frac{6.626×10^{-34}×3×10^8×0.02427×10^{-10}}{1.0872×10^{-10} ×1.09045×10^{-10}} = 4.069×10^{-17} J $ $v = √\frac{2×K.E.}{m}=√\frac{2×4.071×10^{-17}}{9.1×10^{-31}} = 9.45×10^6 m/s$ 20. X-rays of wavelength 4Aº are scattered from a block of carbon. The scattered x-rays are observed at an angle of 45º to the incident beam. Calculate the Compton shift and the fraction of energy lost by the photon in this collision. Solution: λ=4 A° φ = 45° λ' = ? Fraction of energy lost = ? $λ' = λ + \frac{h}{m_0c} (1 - cos φ) = 4 + 0.02427(1 - cos 45) = 4.00710 A°$ Fraction of energy lost = $\frac{E-E'}{E} = \frac{hc}{λ'} - \frac{hc}{λ} = \frac{hc}{λ} * \frac{Δλ}{λ'} $ $Δλ = λ' - λ = 4.00710 - 4 = 0.007104 A°$ $\frac{Δλ}{λ'} = \frac{0.00710}{4.00710} = 0.00177 $ Thus, the fraction of energy lost by the photon in this collision is 0.177 %. ##Engineering Physics B.E. First Semester / 27 21. Using de Broglie hypothesis calculate the wavelength associated with the stone of mass 50 gms moving with speed of 50 m/s and an electron with K.E. of 100 eV. Solution: m=50gms = 0.5 kg v= 50m/s K.E = 100eV $λ_{stone}= ?$, $λ_{e} = ?$ $λ_{stone} = \frac{h}{mv} = \frac{6.626×10^{-34}}{0.5×50}= 2.65×10^{-35} m$ $λ_{e}= \frac{h}{√2mE} = \frac {6.626×10^{-34}}{√2×9.1×10^{-31} ×1.602×10^{-19} ×100} = 1.224 A°$ 22. Calculate the de Broglie wavelength associated with an electron with energy 60 eV. Solution: E = 60 eV, λ = ? $λ_{e} = \frac{h}{√2mE} = \frac {6.626×10^{-34}}{√2×9.1×10^{-31} ×60×1.602×10^{-19}} = 1.5844 A°$ ## Engineering Physics B.E. First Semester / 28 23. Calculate de Broglie wavelength of an electron subjected to a potential difference of 1.5kV. Solution: V=1.5kV, λ = ? $λ_{e} = \frac{h}{√2meV} = \frac {6.626×10^{-34}}{√2×9.1×10^{-31} ×1.602×10^{-19} ×1.5×1000} = 0.317 A°$ 24. Electrons of energy 10 KeV are passed through a thin film of metal for which atomic spacing is 5.5x10^-11 m. What is the angle of deviation of the first order diffraction maxima? Solution: d=5.5x10 ^-11 m, E=10 KeV, θ₁ =? $λ = \frac{h}{√2mE} = \frac {6.626×10^{-34}}{√2×9.1×10^{-31} ×10^4×1.602×10^{-19}} = 1.22×10^{-11} m$ $θ₁ = sin^{-1} \frac{ηλ}{2d} = sin^{-1} \frac{1 × 1.22 × 10^{-11}}{2×5.5 × 10^{-11}} = 6.36 ° $ 25. Electrons of are accelerated from rest through a potential difference of 150 V. These electron waves are diffracted by a grating with a spacing of 2A°. (i) Calculate the de Broglie wavelength of the electron. (ii) At what angle measured from the main axis will the first order maximum be observed? Solution: V = 150V m = 9.1x10^-31 kg d = 2A° = 2x10^-10 m n = 1 λ = ? θ₁ = ? $λ = \frac{h}{√2meV} = \frac {6.626×10^{-34}}{√2×9.1×10^{-31} ×1.602×10^{-19} ×150} = 1.001 × 10^{-10} A°$ $θ₁ = sin^{-1} (\frac{ηλ}{2d}) = sin^{-1} (\frac{1 × 1.001 × 10^{-10}}{2×2 × 10^{-10}}) = 14.50°$ 26. Find the de Broglie wavelength of (i) an electron accelerated through a potential difference of 182 V and (ii)1 Kg object moving with speed of 1 m/sec. Solution: V = 182V, m=1 kg, v=1m/s $λ_{e} = \frac{12.27}{√V} = \frac{12.27}{√182} = 0.9094 A°$ $λ_{object} = \frac{h}{mv} = \frac {6.626×10^{-34}}{1×1} = 6.626 × 10^{-34} m$