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A photosensitive metallic surface has work function, hv0. If photons of energy 2hv0 fall on this surface, the electrons come out with a maximum velocity of 4 x 10^6 m/s. When the photon energy is increased to 5hv0, then maximum velocity of photoelectrons will be :

  1. A
    16 x 10^6 m/s
  2. B
    8 x 10^7 m/s
  3. C
    4 x 10^5 m/s
  4. D
    8 x 10^6 m/s

Solution & Step-by-step Explanation

The kinetic energy (KE) of the photoelectrons emitted is given by KE = hf - φ, where hf is the energy of the photon and φ is the work function of the metal. When the photon energy is 2hv0, the kinetic energy of the electrons is 2hv0 - hv0 = hv0, and this corresponds to a maximum velocity of 4 x 10^6 m/s. The kinetic energy can be expressed as (1/2)mv^2, where m is the mass of the electron. Thus, for the first case, (1/2)mv^2 = hv0. When the photon energy is increased to 5hv0, the kinetic energy of the electrons becomes 5hv0 - hv0 = 4hv0. Since the kinetic energy is now 4 times the initial kinetic energy, the velocity will increase by a factor of √4 = 2. Therefore, the new maximum velocity will be 2 * 4 x 10^6 m/s = 8 x 10^6 m/s.

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A photosensitive metallic surface has work function, hv0. If photons of energy 2hv0 fall on this surface, the electrons come out with a maximum velocity of 4 x 10^6 m/s. When the photon energy is increased to 5hv0, then maximum velocity of photoelectrons will be :
A
16 x 10^6 m/s
B
8 x 10^7 m/s
C
4 x 10^5 m/s
D
8 x 10^6 m/s

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