## 7th Edition 1B.15

$E=hv$

klarratt2
Posts: 60
Joined: Fri Sep 28, 2018 12:16 am

### 7th Edition 1B.15

For part (c), I used the equation c=(wavelength)(frequency) when frequency=2.5x10^16 (as given in the problem), and I got 12nm for the wavelength. However, in the back of the book, it says the solution is 8.8nm. What am I doing wrong?

Francis_Nguyen3D
Posts: 35
Joined: Fri Sep 28, 2018 12:24 am

### Re: 7th Edition 1B.15

For this problem use the equation: Energy= hc/(wavelength).
Then use the energy of the photon you got from part b, which should be 2.25 x 10^-17J.
Rearrange the equation so that you are solving for wavelength so it would be: wavelength= hc/energy
Then plug in 6.626 x 10^-34 for h and 3.0 x 10^8 for c and divide by 2.25 x 10^-17 and you should get around 9nm.
The book gets 8.8nm from using the unrounded numbers for Planck's constant and the speed of light.

Imelda Mena 3I
Posts: 51
Joined: Thu Oct 02, 2014 3:00 am

### Re: 7th Edition 1B.15

I am having the same problem, I am getting 12nm for the wavelength instead of the 8.8nm that the solutions manual says. My answer for part b was 1.66x10^-17J, not 2.25x10^-17J. How did you get 2.25x10^-17J?

Imelda Mena 3I
Posts: 51
Joined: Thu Oct 02, 2014 3:00 am

### Re: 7th Edition 1B.15

I got it now! The energy of the photon that should be used to solve for wavelength in part c is in fact 2.25 x 10^-17J which is obtained from adding up the energy from part b (1.66 x 10^-17J) to the kinetic energy (1/2mv^2) of the electron. Once you do that you should get 2.25 x 10^-17J which is used in the equation $\lambda =hc/E(photon)$ to get a wavelength of 9nm. Hope this helps!