##### physics question 1 hw 20

label Physics
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5

1. When light of wavelength 322 nm falls on a potassium surface, electrons are emitted that have a maximum kinetic energy of 1.35 eV. What is the work function of potassium? The speed of light is 3 × 108 m/s and Planck’s constant is 6.63 × 10−34 J · s. Answer in units of eV.

2. What is the cutoff wavelength of potassium? Answer in units of nm.

3.What is the threshold frequency for potassium? Answer in units of Hz.

Oct 23rd, 2017

Thank you for the opportunity to help you with your question!

$\\ \lambda=322\hspace{5}nm=322\times10^{-9}\hspace{5}nm\\ \\ K_{max}=1.35\hspace{5}eV\\ \\ h=6.63\times10^{-34}\hspace{5}Js=\frac{6.63\times10^{-34}\hspace{5}Js}{1.6\times10^{-19}\hspace{5}J/ev}=4.13\times10^{-15}\hspace{5}eVs\\ \\$

$\lambda=322\hspace{5}nm=322\times10^{-9}\hspace{5}m$

(Please ignore the first step)

According to Einstein's photoelectric equation

$\\ h\upsilon =\phi +K_{max}\\$

where $\\ \phi$ is the work function.

$\\ \phi=h\upsilon-K_{max}\\ \\ \phi=\frac{hc}{\lambda}-K_{max}\\ \\ \phi=\frac{4.13\times10^{-15}\hspace{5}eVs\times3\times10^8\hspace{5}m/s}{322\times10^{-9}\hspace{5}m}-1.35\hspace{5}eV\\ \\ \phi=3.85\hspace{5}eV-1.35\hspace{5}eV\\ \\ \phi=2.5\hspace{5}eV$

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 15th, 2015

Shall submit the remaining answer in 10 minutes.

Jul 15th, 2015

To find the cutoff wavelength, equate the energy of incident light and the work function.

$\\ h\upsilon_{\circ}=\phi\\ \\ \frac{hc}{\lambda_{\circ}}=\phi\\ \\ \lambda_{\circ}=\frac{hc}{\phi}\\ \\ \lambda_{\circ}=\frac{4.13\times10^{-15}\hspace{5}eVs\times3\times10^8\hspace{5}m/s}{2.5\hspace{5}eV}\\ \\ \lambda_{\circ}=4.956\times10^{-7}\hspace{5}m\\ \\ \lambda_{\circ}=495.6\times10^{-9}\hspace{5}m\\ \\ \lambda_{\circ}=495.6\hspace{5}nm\\$

This is the cutoff wavelength.

Threshold frequency is

$\\ \upsilon_{\circ}=\frac{c}{\lambda_{\circ}}\\ \\ \upsilon_{\circ}=\frac{3\times10^8\hspace{5}m/s}{4.956\times10^{-7}\hspace{5}m}\\ \\ \upsilon_{\circ}=6.05\times10^{14}\hspace{5}Hz$

If the frequency of light falls below threshold frequency then no electrons will be emitted.

Jul 15th, 2015

Sorry for the delay.

Jul 15th, 2015

...
Oct 23rd, 2017
...
Oct 23rd, 2017
Oct 24th, 2017
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
Final Answer

Secure Information

Content will be erased after question is completed.

check_circle
Final Answer