Monochromatic light of wavelength ejects photoelectrons from a cesium surface having a work function of = 1.9 eV . These photoelectrons are made to collide with hydrogen atoms initially in the ground state. Calculate the maximum value of so that the excited hydrogen atoms may subsequently emit visible light.

Step-by-step solution

For hydrogen atoms to emit visible light, they must be excited to at least the n=3 state, as visible light corresponds to the Balmer series (transitions to n=2 ). The minimum energy required to excite a hydrogen atom from the ground state ( n=1 ) to n=3 is: E = 13.6 ( 1 1^2 - 1 3^2 ) = 13.6 8 9 = 12.09 eV The photoelectrons must have a maximum kinetic energy K_ max of at least 12.09 eV to cause this excitation. Using the photoelectric equation: K_ max = hc - Substituting the known values and hc 1242 eV nm : 12.09 =