An electron in a hydrogen atom is initially in a state where its de Broglie wavelength is 4 ₀ (where ₀ is the de Broglie wavelength in the ground state). It makes a transition to a lower energy state where its de Broglie wavelength is 2 ₀ . The photon emitted during this transition falls on a metal surface having a work function of 1.05 eV . The stopping potential for the emitted photoelectrons is (Given: Ground stat

Step-by-step solution

The de Broglie wavelength of an electron in the n^ th Bohr orbit is given by = 2 r n . Since the orbital radius r n^2 , we have n . Given that the initial de Broglie wavelength is 4 ₀ and the final is 2 ₀ , the electron transitions from n_i = 4 to n_f = 2 . The energy of the emitted photon is: E = 13.6 ( 1 n_f^2 - 1 n_i^2 ) eV E = 13.6 ( 1 2^2 - 1 4^2 ) = 13.6 ( 1 4 - 1 16 ) = 13.6 3 16 = 2.55 eV According to Einstein's photoelectric equation, the maximum kinetic energy of the photoelectrons is K_ = E - . The stopp