## Ex 1.33 6th edition

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

FrankieClarke2C
Posts: 59
Joined: Fri Sep 28, 2018 12:28 am

### Ex 1.33 6th edition

For part c it asks you to calculate the wavelength of the energy that was used to eject the electron from the metallic surface. I thought you would just take the energy and plug it into the equation, however, they plugged the energy from part b into the kinetic energy equation. They then used this new energy for part c. Why wouldn't you just use the energy originally calculated?

Cody Do 2F
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am

### Re: Ex 1.33 6th edition

For part (a) of question 33, you use the De Broglie equation in order to find the wavelength of the ejected electron. Thus, energy is not calculated in part (a). Part (b) provides to the frequency of light that hits the metal and ejects the electron. Thus you use the frequency to find the energy of this light. Part (c) once again refers to the light that hits the metal (part b) NOT the ejected electron (part a). That's why you use the value calculated in part (b).