(i) Impact parameter (b) is the perpendicular distance of the initial velocity vector of the alpha-particle from the center of the nucleus. Distance of closest approach (r0) is the distance of the alpha-particle from the nucleus at the moment it comes to momentary rest before reversing its direction. (ii) For scattering angle theta = 0, b = infinity (or maximum possible). For theta = 180 degrees, b = 0 (head-on collision).
According to Einstein's photoelectric equation, K = hv - hv0. For frequency v1: K = hv1 - hv0 (1). For frequency v2: 2K = hv2 - hv0 (2). Multiplying (1) by 2: 2K = 2hv1 - 2hv0 (3). Equating (2) and (3): hv2 - hv0 = 2hv1 - 2hv0. This gives hv0 = 2hv1 - hv2. Thus, the threshold frequency v0 = (2v1 - v2).
A Light Emitting Diode (LED) is a heavily doped p-n junction that emits spontaneous radiation when forward-biased. When forward-biased, electrons from the n-region and holes from the p-region move toward the junction and recombine. This recombination of charge carriers releases energy in the form of photons. Uses: (1) In optical communication and (2) In display panels (TVs, smartphones, etc.).
When a p-n junction is formed, the difference in carrier concentration leads to the diffusion of holes from the p-side to the n-side and electrons from the n-side to the p-side. This diffusion leaves behind uncompensated ionized acceptors in the p-region and ionized donors in the n-region near the junction. This space charge region, which is devoid of free mobile charge carriers, is called the depletion region.
A full-wave rectifier uses two diodes. During the first half of the AC cycle, one diode is forward-biased and conducts; during the second half, the other diode is forward-biased. The current through the load resistance flows in the same direction for both halves. [Circuit Diagram: Two diodes D1 and D2 connected to a center-tapped transformer with load RL]. The input waveform is a continuous sine wave, while the output waveform consists of positive half-cycles for both the positive and negative input cycles.
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Log in to view solution →English physicist Thomas Young explained the interference of light using the principle of superposition of waves. He observed the interference pattern on a screen through his experimental arrangement, now called Young's double-slit experiment. He illuminated two slits S1 and S2 with light coming from a slit S. This interference pattern is made of bright and dark bands. These types of bands are called fringes. The distance between two consecutive bright and dark fringes is called fringe width.
- A. decreases, but intensity of bright fringe remains the same
- B. increases, but intensity of bright fringe decreases
- C. decreases, and intensity of bright fringe decreases
- D. and intensity both remain the same
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Log in to view solution →- A. the pattern will remain unchanged
- B. pattern intensity will decrease
- C. fringes will disappear
- D. intensity will increase
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Log in to view solution →The British physicist Thomas Young explained the interference of light using the principle of superposition of waves. He observed the interference pattern on the screen, in his experimental set-up, known as Young's double slit experiment. The two slits S1 and S2 were illuminated by light from a slit S. The interference pattern consists of dark and bright bands of light. Such bands are called fringes. The distance between two consecutive bright and dark fringes is called fringe width.
- A. will decrease, but the intensity of bright fringe remains the same.
- B. will increase, but the intensity of bright fringe decreases.
- C. will decrease, but the intensity of bright fringe increases.
- D. and the intensity both remain the same.
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Log in to view solution →- A. The intensity of pattern will increase
- B. The intensity of pattern will decrease
- C. The number of fringes will become double
- D. No pattern will be observed on the screen
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Log in to view solution →- A. same amplitude and have a varying phase difference.
- B. same wavelength and a constant phase difference.
- C. different wavelengths and same intensity.
- D. different wavelengths and a constant phase difference.
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Log in to view solution →- A. beta/n
- B. beta * n
- C. beta
- D. 2 * beta
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Log in to view solution →- A. bright fringes are formed at both points.
- B. dark fringes are formed at both points.
- C. a bright fringe is formed at P1 and a dark fringe is formed at P2.
- D. a bright fringe is formed at P2 and a dark fringe is formed at P1.
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