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JEE MainPhysicsDual Nature of Matter

Radiation emitted by a hydrogen atom undergoing a transition from the state n=3 to n=2 falls on two different metals P and Q . The work functions of metals P and Q are 1.5 eV and 2.5 eV respectively. Choose the correct statement regarding the emission of photoelectrons.

Options

  1. AOnly metal Q will emit photoelectrons
  2. BOnly metal P will emit photoelectrons
  3. CBoth metals P and Q will emit photoelectrons
  4. DNeither metal P nor metal Q will emit photoelectrons

Correct answer

B. Only metal P will emit photoelectrons

Step-by-step solution

The energy of the photon emitted during the transition from n=3 to n=2 in a hydrogen atom is given by: E = 13.6 ( 1 2^2 - 1 3^2 ) eV E = 13.6 ( 1 4 - 1 9 ) eV = 13.6 5 36 eV 1.89 eV For photoelectric emission to take place, the energy of the incident photon must be greater than or equal to the work function of the metal ( E ). For metal P : E = 1.89 eV > 1.5 eV . Thus, metal P will emit photoelectrons. For metal Q : E = 1.89 eV Therefore, only metal P will emit photoelectrons. Answer: Only metal P will emit photoel

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