1.31

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ChristianM3F
Posts: 61
Joined: Sat Sep 14, 2019 12:16 am

1.31

Postby ChristianM3F » Mon Oct 28, 2019 10:41 pm

Can someone help me understand the solution for 1.31? I have no idea where they got the equation/values used to calculate E (work function).
The only way I know how to calculate work function is using the amount of light energy and the kinetic energy of the electron that's ejected, so I'm majorly confused.

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

Re: 1.31

Postby nicolely2F » Mon Oct 28, 2019 10:45 pm

The value of lithium's work function is given at the end of the exercise. It's the last sentence in the prompt, so a bit easy to miss.

Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am

Re: 1.31

Postby Chris Tai 1B » Sat Nov 02, 2019 5:42 pm

The question is a bit tricky because it also doesn't give the work function in terms of joules, which is what you need to plug into the equation E(photon) = E(work function)+E(kinetic energy of e-). Rather, it's given in electronvolts: 2.93 electron volts. There are 1.60218x10^-19 J in 1 eV, so we convert by multiplying 2.93 * 1.60218x10^-19 to get 4.6944x10^-19 J. Through this, we can find that only the violet GaN laser will emit electrons, due to its higher frequency, (4.91x10^-19 > 4.69x10^-19) thus enabling the sensor to remain activated.


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