Question

a) Electrons are ejected from a photoelectric surface with a maximum speed of 4.20 x 10⁵ m/s. If the work function of the metal is 2.55 eV, what is the wavelength of the incident light ? Express your answer in the format of a.bc x 10^-d m

b) A photoelectric cell is illuminated with white light (wavelengths from 400 nm to 700 nm). What is the maximum kinetic energy (in eV) of the electrons emitted by this surface if its work function is 2.30 eV ? Input your 4 digit answer

c) An electron in the 3^rd stationary state around a hydrogen atom has an energy of -1.512 eV. What will be the electron's transition energy be (in eV) if the hydrogen atom emits a photon with a wavelength of 6.57 x 10^-6 m ? Express your 3 digit answer

d) What is the shortest wavelength of photon that can be emitted in an hydrogen atom ? Express your answer in the format of a.bc x 10^-d m

Answer 1

Multiple questions posted, so answering only the first question:

a) For photoelectric effect, hc/=+K, where h is plank's constant=6.626*10^-34 m2-kg/s, is wavelength of incident light,c is speed of light, is work function and K is maximum kinetic energy.

For the given problem, =2.55 eV=2.55*1.6*10^-19 J = 4.08*10^-19 J

Also, K=1/2*m*v2, where m is mass of electron and v is its maximum speed.Here,v=4.2*10^5 m/s.

So, K=1/2*(9.11*10^-31)*(4.2*10^5)^2 = 8.035*10^-20 J.

Hence, hc/=4.08*10^-19+8.035*10^-20 = 4.8835*10^-19

=>=hc/(4.8835*10^-19 ) = (6.626*10^-34)*(3*10^8)/(4.8835*10^-19 )

=>=4.07*10^-7 m.