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Ashok0210

Asked: April 27, 2021In: Class 12 Physics

An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (∼10⁻² mm of Hg). A magnetic field of 2.83 × 10⁻⁴ T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data

Class 12 Physics CBSE and UP Board Dual Nature of Radiation and Matter Chapter-11 Exercise 11.22 NCERT Solutions for Class 12 Physics Chapter 11 Question-22 Additional Exercise

Ashok0210

Asked: April 27, 2021In: Class 12 Physics

(a) A monoenergetic electron beam with electron speed of 5.20 × 10⁶ m s⁻¹ is subject to a magnetic field of 1.30 × 10⁻⁴ T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 10¹¹C kg⁻¹. (b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified? [Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]

Class 12 Physics CBSE and UP Board Dual Nature of Radiation and Matter Chapter-11 Exercise 11.21 NCERT Solutions for Class 12 Physics Chapter 11 Question-21 Additional Exercise

Ashok0210

Asked: April 27, 2021In: Class 12 Physics

(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 10¹¹ C kg⁻¹. (b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong ? In what way is the formula to be modified?

Class 12 Physics CBSE and UP Board Dual Nature of Radiation and Matter Chapter-11 Exercise 11.20 NCERT Solutions for Class 12 Physics Chapter 11 Question-20 Additional Exercise

Ashok0210

Asked: April 27, 2021In: Class 12 Physics

What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

Class 12 Physics CBSE and UP Board Dual Nature of Radiation and Matter Chapter-11 Exercise 11.19 NCERT Solutions for Class 12 Physics Chapter 11 Question-19

Ashok0210

Asked: April 27, 2021In: Class 12 Physics

Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Class 12 Physics CBSE and UP Board Dual Nature of Radiation and Matter Chapter-11 Exercise 11.18 NCERT Solutions for Class 12 Physics Chapter 11 Question-18

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What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

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