Physics1

1.) If binding energy per nucleon in Ne²⁰ , He⁴ and C¹²  nuclei are equal to 8.03, 7.07 and 7.68 MeV respectively then energy required for a Ne²⁰ nucleus into two α –particles and a C¹² nucleus is

     (A) 11.9 MeV

     (B) 1.9 MeV

     (C) 12.5 MeV

     (D) 20 MeV

2.) In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 10¹⁰ Hz and its amplitude is 48 Vm^-1 . The wavelength of the wave is

     (A) 1.5 m

     (B) 1.5 × 10^-3 m

     (C) ) 1.5 × 10^-2 m

     (D) 24 × 10¹⁰ m

3.) A load resistor of 2 kΩ is connected in the collector branch of an amplifier circuit using a transistor in common-emitter mode.The current gain β= 50. The input resistance of a transistor is 0.50 kΩ. If the input current is changed by 50 µA then change in Output voltage is

   (A) 0.5V

   (B) 5V

   (C) 2V

   (D) 20V

4.) If radiation corresponding to the transition n=4 to n=2 from Hydrogen atoms falls on Cesium metal( work function=1.9eV). The maximum kinetic energy of the emitted electrons is

    (A) 1eV

    (B) 0.065eV

    (C) 0.5eV

    (D) 0.65eV

5.) A beam of light of wavelength 600nm from a distant source falls on a single slit 1mm wide and resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes one either side of the central bright fringe is

    (A) 1.2 cm

    (B) 1.2 mm

    (C) 2.4 cm

    (D) 2.4 mm

Answer-  1.) A        2.)  C      3.) B        4.) D       5.) D

Hints & Solution

1.) E = [20 × 8.03 – (2 × 4 × 7.07 + 12 × 7.68)] MeV

        = 11.9 MeV

2.)  c = fλ

       λ = 3 × 10⁸ / 2 × 10¹⁰

          = 1.5 × 10^-2 m

4.) Energy of photon emitted during transition from n = 4 to n = 2

     = 13.6[ (1/4) – (1/16) ] = 2.55eV

     Therefore Kmax = 2.55 – 1.9 = 0.65eV

5.) Distance between 1^st dark fringes on either side of central bright fringe

     = width of central bright fringe = 2λD/d

    λ = 600 × 10^-9 m, D = 2m, d= 10^-3 m