1B #15

$\lambda=\frac{h}{p}$

dtolentino1E
Posts: 101
Joined: Thu Jul 11, 2019 12:17 am

1B #15

hi! would someone please walk me through their steps for this problem?

The velocity of an electron that is emitted from a metallic surface by a photon is 3.63 e 3 km.s^-1. (a) What is the wavelength of the ejected electron? (b) No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50 3 e 16 Hz. How much energy is required to remove the electron from the metal surface? (c) What is the wavelength of the radiation that caused photo ejection of the electron? (d) What kind of electromagnetic radiation was used?

Posts: 54
Joined: Sat Jul 20, 2019 12:17 am

Re: 1B #15

For part A we know the de broglie equation is lambda=h/mv. The question asks us to solve for wavelength, planck's constant is always the same as well as the mass of an electron, and the velocity is given in the question so the answer can be found simply by plugging in everything we know and carrying out the algebra to the final answer.

Posts: 54
Joined: Sat Jul 20, 2019 12:17 am

Re: 1B #15

For part B we need the equation which gives the energy of a photon, E=hv (note that v is not actually v it is another variable that stands for frequency). h is planck's constant which we already know, E is the energy which we are solving for, and the frequency at which an electron is drawn off the surface is given in the problem therefore we can again just plug in the values we know and carry out the algebra until the answer is reached simply by multiplying h by the frequency given.

Posts: 54
Joined: Sat Jul 20, 2019 12:17 am

Re: 1B #15

For part c we need a simple equation involving the speed of light which is c=frequency times lambda. The speed of light is a constant we know, frequency was given earlier in the problem, hence we can just plug in the information we know again and carry out the algebra to solve for lambda which is the wavelength we're looking for.