Respuesta :
Answer:
The maximum wavelength of second laser is 453.8 nm
Explanation:
Given:
Wavelength of light [tex]\lambda = 770 \times 10^{-9}[/tex] m
Ionization energy of potassium [tex]= 6.96 \times 10^{-19}[/tex] J
First find the energy related to the one of light,
[tex]E = \frac{hc}{\lambda}[/tex]
Where [tex]h = 6.626 \times 10^{-34}[/tex], [tex]c = 3 \times 10^{8} \frac{m}{s}[/tex]
[tex]E = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8} }{770 \times 10^{-9} }[/tex]
[tex]E = 2.58 \times 10^{-19}[/tex]J
This is the energy of the first laser, the missing energy is given by.
[tex]E' = (6.96 - 2.58) \times 10^{-19}[/tex]
[tex]E ' = 4.38 \times 10^{-19}[/tex]J
For finding wavelength of second laser,
[tex]E' = \frac{hc}{\lambda '}[/tex]
[tex]\lambda ' = \frac{hc}{E'}[/tex]
[tex]\lambda ' = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8} }{4.38 \times 10^{-19} }[/tex]
[tex]\lambda ' = 453.8 \times 10^{-9 }[/tex] m
[tex]\lambda ' = 453.8[/tex] nm
Therefore, the maximum wavelength of second laser is 453.8 nm
