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
- AOnly metal Q will emit photoelectrons
- BOnly metal P will emit photoelectrons
- CBoth metals P and Q will emit photoelectrons
- 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