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BITSAT2012PhysicsDual Nature of MatterActual

In an experiment on photoelectric effect photons of wavelength 300 nm eject electrons from a metal of work function 2.25 eV . A photon of energy equal to that of the most energetic electron corresponds to the following transition in the hydrogen atom:

Options

  1. An=2 to n=1 state
  2. Bn=3 to n=1 state
  3. Cn=3 to n=2 state
  4. Dn=4 to n=3 state

Correct answer

C. n=3 to n=2 state

Step-by-step solution

Given =2.25 eV =300 100⁻³ ~m Energy of the given photon, E = hv = hc = 663 10⁻³⁴ 3 10^8 300 10⁻⁹ =6.63 10⁻¹⁹ ~J =4.13 eV Now, E = hc - =4.137-2.25=1.88 eV (1) So, 1.88 eV energy is used to jump from one orbit to another orbit by electron. Therefore, energy of different orbital of hydrogen array rrrr n =1 & 2 & 3 & 4 E =-13.6 ev & -3.4 & -1.51 & -0.85 array And from statement (1), 1.18=E_i-E_f E_i & E_f . are energy of initial and final orbit And also from the above table, we can observe that, E_ i - E _ f =(-1.51)-

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