EXERCISE SHORT QUESTIONS
CHAPTER # 19: DAWN OF MODERN PHYSICS
Q # 1. What are the measurements on which two observers in relative motion will always agree upon?
Ans. The measurement on which two observers in relative motion will always agree upon is speed of light.
Q # 2. Does the dilation means that time really passes more slowing in moving system or that it only seems to pass more slowly?
Ans. According to the time dilation formula \(= \frac{t_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}\) , time is not constant. It is relative.
Time passes normally for any observer within his own system.
Time seems to pass more slowly when an observer in one system in relativistic motion takes the time measurement of the other system.
Q # 3. If you are moving in a space ship at very high speed relative to the earth, would you notice a difference (a) in your pulse rate (b) in pulse rate of people on earth?
Ans. The pulse rate of a person who is travelling in a spaceship is not changed with respect to clock inside the spaceship. But the person in spaceship will experience the change in pulse rate of the people on earth, according to the relation \(t = \frac{t_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}\).
Q # 4. If the speed of light were infinite, what would the equations of special theory of relativity reduce to?
Ans. If we take speed of light c as infinity, then the equations of special theory of relativity reduce to:
Time dilation formula: \(t = \frac{t_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} = \frac{t_{0}}{\sqrt{1 - \frac{v^{2}}{\infty^{2}}}} = \frac{t_{0}}{\sqrt{1 - 0}} = t_{0}\), i.e., Time in motion=Proper Time
Length contraction formula: \(L = L_{0}\sqrt{1 - \frac{v^{2}}{c^{2}}} = L_{0}\sqrt{1 - \frac{v^{2}}{\infty^{2}}} = L_{0}\sqrt{1 - 0} = L_{0}\), i.e., Length in motion = Proper Length
Mass increment formula: \(m = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} = \frac{m_{0}}{\sqrt{1 - \frac{v^{2}}{\infty^{2}}}} = \frac{m_{0}}{\sqrt{1 - 0}} = m_{0}\), i.e., Mass in motion=Rest Mass
Q # 5. Since mass is form of energy, can we conclude that a compressed spring has more mass than the same spring when it is not compressed?
Ans. According to the theory of relativity,Mass is form of energy. As compressed spring has greater potential energy, so there would be increase in mass of compressed spring according to the relation: \(\mathrm{\Delta}m = \frac{\mathrm{\Delta}E}{c^{2}}\). However, this increase in mass is slightly greater than original mass (negligibly small).
Q # 6. As a solid is heated and begins to glow, why does it first appear red.
Ans. At lower temperature, a body emits radiation of low energy (longer wavelength). Since longest visible wavelength is red, so it appears red first.
Q # 7. What happens to total radiation from a black body if its absolute temperature is doubled?
Ans. According to Stephen Boltzmann law: \(E = \sigma T^{4}\)
When absolute temperature is doubled, then: \(E^{'} = \sigma(2T)^{4} = 16\sigma T^{4} = 16E\)
Thus if absolute temperature is doubled, the total radiation emitted by black body increases 16 times.
Q # 8. Which photon, red, green or blue carry the most (a) energy and (b) momentum?
Energy: According to relation: \(E = hf = \frac{hc}{\lambda}\), the photons of blue light having shorter wavelength must have larger energy as compared to photons of red or green color light.
Momentum: According to relation: \(p = \frac{h}{\lambda}\), the photons of blue light having shorter wavelength must have larger momentum as compared to photons of red or green color light.
Q # 9. Which have the low energy quanta? Radio waves or X-rays.
Ans. According to relation: \(E = hf = \frac{hc}{\lambda}\), the quanta of X-rays having shorter wavelength must have larger energy as compared to quanta of radio waves.
Q # 10. A beam of red light and a beam of blue light have exactly the same energy. Which beam contains the greater number of photons?
Ans. As
(Energy of a photon of blue light) > (Energy of a photon of red light),
Therefore, two color beams having same energy will contain different number of photons.
The blue light, having photon of comparatively larger energy contains less number of photons
The red light, having photon of comparatively smaller energy contains greater number of photons
Q # 11. Does the brightness of a beam of light primarily depend on the frequency of photons or the number of photons?
Ans. The brightness of a beam depends upon intensity (number of photons) and not on the frequency of light. Thus brightness increases with intensity of light.
Q # 12. When ultravoilot light falls on certain dyes, visible light is emitted. Why does this not happens when infrared light falls on these dyes?
Ans. UV light consists of photons having energy greater than energy of visible light photons. When UV light falls on dyes, atoms initially become excited and then de-excited by emitting lower energy photons, which may be detectable by normal human eyes.
Infrared light consists of photons having energy lower than energy of visible light photons. When Infrared light falls on dyes, atoms initially become excited and then de-excited by emitting lower energy photons which couldn't lie in visible spectrum of electromagnetic radiation.
Q # 13. Will bright light eject more electrons from metal surface than dimmer light of same color?
Ans. Since ``number of electrons'' ejected from metal surface depend upon the intensity of light (number of photons). Therefore, bright light being more intense will eject more electrons from a metal surface than dimmer light of same color.
Q # 14. Will higher frequency light eject greater number of electrons than lower frequency light?
Ans. No, the higher frequency light will not eject greater number of electrons than low frequency light. It is because of the reason that number of electrons emitted from metal surface depends upon intensity of light (number of photons) and not frequency of light.
Q # 15. When light shines on a surface, is momentum transferred to the metal surface?
Ans. When light falls on the surface, about 20\% of incident light energy is absorbed in each reflection. So both energy and momentum is transferred to the metal surface.
Q # 16. Why can red light be used in photographic dark room when developing films but a blue or white light cannot?
Ans. Since the frequency of red light is less as compared to blue light, so red light has less energy as compared to blue light. Therefore, photographic films an the material concerned are less affected in the presence of red light.
Q # 17. Photon A has twice the energy of photon B. what is the ratio of the momentum to A to that of B.
Ans. Given that the energy of photon A is twice the energy of photon B i.e.,
\[E_{A} = 2E_{B}\]
The momentum of photon A = \(P_{A} = \frac{E_{A}}{c}\)
The momentum of photon B = \(P_{B} = \frac{E_{B}}{c}\)
Now,
\(\frac{P_{A}}{P_{B}} = \frac{\left( \frac{E_{A}}{c} \right)}{\left( \frac{E_{B}}{c} \right)} = \frac{E_{A}}{E_{B}} = \frac{2E_{B}}{E_{B}} = 2\)
So, photon A has twice the momentum of photon B.
Q # 18. Why don't we observe Compton effect with visible light?
Ans. We don't observe a Compton effect with visible light because photons of visible light have smaller energy and momentum then the photons of X-rays.
Q # 19. Can pair production takes place in vaccum? Explain.
Ans. No, pair production can't take place in vacuum. Because, in vacuum, there is no heavy nucleus present. Pair production always takes place in the presence of a heavy nucleus.
Q # 20. Is it possible to create a single electron from energy? Explain.
Ans. No it is not possible to create a single electron from energy. The creation of single electron from energy is violation of law of conservation of charge. Whenever pair production takes place, the electrons and positrons are created at the same time.
Q # 21. If electrons behaved only light particles, what pattern would you expect on the sucreen after the electron passing through double slit?
Ans. If electron behave only like particles then, after passing through the double slit, only those parts of the screen are affected which are in front of the slits.
Q # 22. If an electron and proton have the same de Broglie wavelength, which particle have greater speed?
Ans. The de Broglie wavelength associated with moving particle is given by expression:
\[\lambda = \frac{h}{mv} \Longrightarrow v = \frac{h}{m\lambda}\]
As the wavelength is same for both electron and proton beam, therefore:\(v \propto \frac{1}{m}\)
As mass of electron is smaller then proton, so electron has greater speed.
Q # 23. We don't notice the de Broglie wavelength for a pitched cricket. Explain. Why?
Ans. The de Broglie wavelength associated with moving particle is given by expression:
\[\lambda = \frac{h}{mv}\]
Due to large mass and small speed, the wavelength associated with moving cricket ball is very small. As the diffraction produced by the ball is also very small. So it is impossible to measure de Broglie wavelength for a pitched cricket ball.
Q # 24. If the following particles all have the same energy, which has the shortest wavelengths? Electrons, \(\mathbf{\alpha -}\)\textbf{particle, neutron, proton.}
Ans. The de Broglie wavelength associated with moving particle is given by expression:
\[\lambda = \frac{h}{mv} = \frac{h}{m\sqrt{\frac{2E}{m}}} = \frac{h}{\sqrt{2mE}}\ \ \ \ \ \ \ \ \ \because E = \frac{1}{2}mv^{2}\]
For same energy of beam of particles, wehave:\(\lambda \propto \frac{1}{\sqrt{m}}\)
Thus the massive particle has shorter wavelength. As mass of alpha particle is greater, so it has the shorter wavelength.
Q # 25.When does light behave as a wave? When does it behave as a particle?
Ans. Light behaves as wave in the phenomenon of:
(i) Interference, (ii) Diffraction, (iii) Polarization
Light behaves as particle in
(i) Photo electric effect, (ii) Compton effect, (iii) Pair production
Q # 26.What advantage an electron microscope has over an optical microscope?
Ans. The resolving power of electron microscope is thousand times greater then an Optical microscope.
The internal structure of an object can also be obtained by electron microscope which is not possible with optical microscope.
Q # 27.If measurement shows a precise position for an electron, can those measurements show precise momentum also? Explain.
Ans. According to Heisenberg's uncertainty principle, it is impossible that measure both position and momentum precisely at the same time. Mathematically:
\[\mathrm{\Delta}x\mathrm{\Delta}p = h\]
Thus if one measurement is precise, then the other is uncertain.
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